Well, let me be self-contained. If, say, $a=b$, then $a(c-1)$ and $a^2-c=2^n$ are powers of 2, hence $a=2^k$, $c-1=2^m$. Considering equality $2^{2k}=a^2=1+(c-1)+(a^2-c)=1+2^m+2^n$ modulo 4 we get $\{m,n\}=\{0,1\}$, $k=1$. This leads to solutions $(2,2,2)$ and $(2,2,3)$. Now assume that $a< b< c$. Denote $bc-a=2^A$, $ac-b=2^B$, then $2^A-2^B=(c+1)(b-a)>0$, hence $A>B$ and $2^B$ divides $(c+1)(b-a)$. Also $2^B$ divides $2^A+2^B=(c-1)(a+b)$. Either $c-1$ or $c+1$ is not divisible by 4, hence either $2(b-a)$ or $2(b+a)$ is divisible by $2^B$, in both cases we have $a+b\geq 2^{B-1}$. So $a(b+1)\leq ac=2^B+b\leq 2(a+b)+b$, i.e. $ab\leq a+3b<4b$, hence $a<4$. If $a=2$, then $2^C=2b-c$, $2^B=2c-b>1$, hence $B>0$. We get $b=(2^{C+1}+2^B)/3$, $c=(2^{B+1}+2^C)/3$. If $C\geq 1$ then both $b,c$ are even, hence $bc-2$ is not divisible by 4, in this case $bc-2\geq 3\cdot 4-2=10$ can not be power of 2. If $C=0$, then by modulo 3 we see that $B$ is even and $bc-2=\frac{(2+2^B)(2^{B+1}+1)-18}9=\frac{2^{2B+1}+5\cdot 2^B-16}{9}$ must be a power of 2. This is not possible if $B\geq 6$, since 32 does not divide numerator $2^{2B+1}+5\cdot 2^{B}-16$ and it is greater then $16\cdot 9$. For $B=2$ we get $b=a=2$, for $B=4$ we get $b=6$, $c=11$, this is another solution. If $a=3$, then $2^C=3b-c$, $2^B=3c-b\geq 2c+1>4$, $B\geq 3$. We get $b=(3\cdot 2^C+2^B)/8$, $c=(3\cdot 2^B+2^C)/8$. We get $B>C$ (from $b<c$) and, since $B\geq 3$, we get that $C\geq 3$ aswell and $2^A=bc-3=(3\cdot 2^{C-3}+2^{B-3})(3\cdot 2^{B-3}+2^{C-3})-3$. If $B>C>3$ this is odd and greater then 1, hence not power of 2. If $C=3$ we get $2^A=2^{B-3}(10+3\cdot 2^{B-3})$. This is a power of 2 if and only if $B-3=1$, $B=4$, $b=5$, $c=7$. This is another solution.

If $(a, b, c)$ is a solution, then $(a, c, b)$, $(b, a, c)$, $(b, c, a)$, $(c, a, b)$, $(c, b, a)$ are also solutions. So assume $WLOG$ that $a\geq b\geq c$. Case $1$: If two numbers are odd and the other is even, then we must have $ab-c=bc-a=1\Rightarrow a=c\Rightarrow a=b=c$, which is impossible. Case $2$: case $2a)$ $a$, $b$ are even and $c$ is odd $\Rightarrow ab-c=1\Rightarrow c=ab-1\geq 2a-1>a$, which is impossible. Case $2b)$ $c$, $a$ are even and $b$ is odd $\Rightarrow ac-b=1\Rightarrow b=ac-1\geq 2a-1>a$, which is impossible. Let case $2c)$ $b$, $c$ are even and $a$ is odd; case $3)$ $a$, $b$, $c$ are all odd; case $4)$ $a$, $b$, $c$ are all even. From $ab-c=2^{\alpha}$ and $ac-b=2^{\beta}$, we obtain $(*)$ $(b-c)(b+c)=2^{\alpha}c-2^{\beta}b$. Observe that $gcd(b, c)$ divides $2^{\beta}\Rightarrow gcd(b, c)=2^{k}$, $k\leq\beta$ and $b=2^{k}b_{1}$, $c=2^{k}c_{1}$, $gcd(b_{1}, c_{1})=1$. From $(*)$, we deduce that $2^{\beta+k}$ divides $2^{2k}(b_{1}-c_{1})(b_{1}+c_{1})\Leftrightarrow$ $(**)$ $2^{\beta-k}$ divides $(b_{1}-c_{1})(b_{1}+c_{1})$. $b=c$ gives $(2, 2, 2)$ and $(3, 2, 2)$, so consider only $b>c\Leftrightarrow b_{1}>c_{1}$. If $\beta=k$, then $ac_{1}-b_{1}=1\Rightarrow b_{1}=ac_{1}-1\geq bc_{1}-1\geq2^{k}b_{1}c_{1}-1\Rightarrow1\geq b_{1}(2^{k}c_{1}-1)\Rightarrow b_{1}=c_{1}=k=1\Rightarrow (2, 2, 2)$ and $(3, 2, 2)$ ($k=0$ and $c_{1}=1$ doesn't give a solution because $c=1$ contradicts $bc-a=2^{\gamma}$), so assume that $k+1\leq\beta$ and observe that $(**)$ now implies that $b_{1}$, $c_{1}$ are both odd. From $(**)$ and the fact that $gcd(b_{1}-c_{1}, b_{1}+c_{1})=2$, we must have $\frac{ac_{1}-b_{1}}{2}=2^{\beta-k-1}\leq b_{1}+c_{1}\Leftrightarrow3b_{1}\geq c_{1}(a-2)$. Case $2c$: $a=bc-1$. If $c_{1}\geq3\Rightarrow b_{1}\geq bc-3\geq 4b_{1}-3\Rightarrow 1\geq b_{1}$, which contradicts $b\geq c$. Hence $c_{1}=1\Rightarrow 3b_{1}\geq bc-3\geq 4b_{1}-3\Rightarrow b_{1}\leq3\Rightarrow b_{1}=3$. $12=3b_{1}+3\geq bc=3\times2^{2k}\Rightarrow k=1\Rightarrow (11, 6, 2)$. Case $3$: Since $b$, $c$ are odd, $gcd(b, c)=2^{k}=1\Rightarrow b=b_{1}, c=c_{1}$. If $c\geq3$, then $3b\geq3(a-2)\Rightarrow b=a-2$ (the case $b=a$ forces $a=1<c$, which is a contradiction). $b=a-2$ leads to $2a+2=2^{\beta}$ and $2a-6=2^{\gamma}$, so $8=2^{\gamma}(2^{\beta-\gamma}-1)\Rightarrow\gamma=3$, $\beta=4$, so $(7, 5, 3)$. Case $4$: If $c_{1}\geq3$, then $3b_{1}\geq3(a-2)\Rightarrow b_{1}\geq a-2\geq b-2\geq 2b_{1}-2\Rightarrow2\geq b_{1}$, which contradicts $b>c$. Hence, $c_{1}=1\Rightarrow ab_{1}-c_{1}=2^{\alpha-k}\Leftrightarrow ab_{1}=2\Rightarrow (2, 2, 2)$.

If $a$ is the greatest number and $ v_2(a)+v_2(b) \neq v_2(c) $ then $ v_2(ab-c) \le v_2(c) \Rightarrow ab-c \le c$ and so $ab \le 2c$. Since $a \ge c$, this implies $b=2$ (since $b=1$ is readily seen to be impossible). But then $2a \le 2c$ and so $a=c$ since $a$ is the greatest. Then $a,a^2-2$ are powers of 2 so $a^2,a^2-2$ are powers of 2, and this leads to the solution $(2,2,2)$. Otherwise, $v_2(a)+v_2(b) = v_2(c), v_2(a)+v_2(c) = v_2(b)$ and so $a$ is odd and $v_2(b)=v_2(c) := d$. If $d>0$, then $bc-a$ is odd so $a=bc-1$ and so $bc^2-b-c, b^2c-b-c$ are powers of 2. If $v_2(b+c) \neq 3d$, then $v_2(bc^2-b-c) = \text{min} \{ 3d, v_2(b+c) \} = v_2(b^2c-b-c)$ and so $bc^2-b-c = b^2c-b-c$, so $b=c$ which leads to the solution $(2,2,3)$. Otherwise, $v_2(b+c)=3d$. But if $b > c$ and $b^2c-b-c = 2^x > 2^y = bc^2-b-c$. So $y = v_2(2^x-2^y) = v_2(b^c-bc^2) = v_2(bc(b-c)) = 3d+1$, since $v_2(b-c)=d+1$ is quickly verified since $v_2(b+c) = 3d > d+1$. So $bc^2 = 2^{3d+1} +b+c$, so if $bc^2=2^{3d}h$, with $h$ odd, then $b+c = 2^{3d}(h-2)$, and so $bc^2 \le 3(b+c)$. This implies $bc^2 \le 6b$ so $c=2$ since $c$ is even. Then $4b \le 3b+6$ so $b=6$ since $b \neq c$ satisfies $v_2(b) = v_2(c)$. Finally, since $a=bc-1$, this leads to the solution $(2,6,11)$. So the final case is when $d=0$, that is, when $a,b,c$ are odd. Let $a \ge b \ge c$, then $2^{\beta} = ca-b \le ab-c$, and $2^{\beta}$ divides both of them. So by $\text{mod} 2^{\beta}$, we get $ab \equiv c, ca \equiv b$, and so $a^b \equiv b$. Since $b$ is odd, $2^{\beta} | a^2-1 = (a-1)(a+1)$. But one of those numbers is $ 2 \text{mod} 4$, and so $2^{\beta-1} = \frac{ca-b}{2}$ divides the other one. This implies $(ca-b)/2 \le a+1$. So $ca-b \le 2a+2$, so $a(c-2) \le b+2$. Since $c$ is odd and $c=1$ is quickly verified to be impossible, we get $c-2 \ge 1$ and so $a \le b+2$. Since $c>1$, we get $a,b >1$ and so $a(c-2) \le b+2 < 2b \le 2a$,and so $c \le 4$ so $c=3$ since $c$ is odd. Since $a \le b+2$, $a=b$ or $b+2$. If $a=b$ then $a^2-3, 3a-a$ are powers of 2, so $2a$ is a power of 2 and so $a=1$ since it's odd. But $1^2-3$ is not a power of 2. So $a=b+2$, and $2b+6, 2b-2$ are powers of 2. So $b+3, b-1$ are powers of 2 and so $b-1 = 4$, so $b=5$ and we get the final solution, $(3,5,7)$.

JuanOrtiz wrote: Disappointed that my last IMO's p2 was a bad problem :/ At least it's an actual number theory problem First one in my recent memory...

N.B : (Cases where the variables have specific values can be easily done , so I just take care of the general case , and skip things like $x \leq 2$ ) First the case where the numbers have different parity , is quite easy to deal with . So I'll just take care of the case where all of them are either odd or even . If all the numbers are even , put $A=v_2(a)>=1$.. , also , suppose wlog that $x>y>z$ Now we have $(a-1)(b+c)(b-1)(a+c)(c-1)(a+b)=(2^x+2^y)(2^y+2^z)(2^z+2^x)$ So $v_2((a+b)(b+c)(c+a))=2z+y$ Now , $2^{2z+y}=(ab-c)^2(ac-b)$ So $v_2((ab-c)^2(ac-b))=2z+y=v_2(a+b)(b+c)(c+a)$ that is $2v_2(ab-c)+v_2(ac-b)=v_2(a+b)+v_2(a+c)+v_2(b+c)$ Then , casework ; Cases where $A \geq B \geq C$ , $C \geq B \geq A$ , $B \geq A \geq C$ implie respectively , $Y=B$ , $Z=A$ , and $Y=A$ all of them can be take care off in the same way ; If $Z=A$ for example , then $a>=2^A.3=3.2^z=3(ab-c)$ , so $c \geq a(b-\frac{1}{3}) \geq ab$ , wich is impossible since $ab-c=2^z> 0$ . Same goes for the others . Now cases where $C \geq A \geq B$ , $A \geq C \geq B$ or $B \geq C \geq A$ are a bit more hairy . If $B \geq C \geq A$ , we split it again into two cases , if $B \geq A+C$ then $2v_2(ab-c)=A \geq C \geq B$ and so $(ab-c)^2|a ,c$ and $ b$ , wich implies $(ab-c)^2|ab-c$ wich cannot hold if ab-c \geq 2 . If $A+C \geq B$ , then $A+B \geq 2B-C\geq2C-C=C$ , implying from the valuations equality that $2B+A=0$ , contradiction. Same fashion applies to other cases . So we're only left with the case where all the numbers are odd : We then put $a=2p+1 , b=2q+1 , c=2r+1$ , working mod 2 in $bc-a=2^x$ and similars implies that either two of $p,q,r$ are odd , or none of them is . In both cases :[wlog $p,r$ are odd and q is even , or all of them are even.] using $(a+c)(b-1)=2^x+2^z$ gives (assuming $x>2$) $ (p+r+1)q=2^{x-2}+2^{z-2}$ , so $q=2^{z-2}$ , that is $ b=2^{z-1}+1 $ , then we have $a+c=2^{x-z+1}+2$ , but $(b+1)(a-c)=2^x-2^z$ gives $a-c=2^{x-z+1}-2$ , sum up and divide by $ 2 $to get $a=2^{x-z+1}$ , contradiction since $a$ is odd. Now dealing with the partiular cases along gives the solutions.

WLOG $ a \geq b \geq c > 1 \Longrightarrow ab-c \geq ca-b \geq bc-a $ $ \boxed{ \text{If a is even.} } $ Since $ 2a-b \leq ca-b = \gcd (ab-c, ca-b) \leq \gcd (ab-c, a(ca-b)+ab-c) \leq c $, so $ a=b=c=2 $ . $ \boxed{ \text{ If a, b, c are odd. } } \Longrightarrow a>b>c>1 $ Since $ ca-b=\gcd(ab-c,ca-b) \leq \gcd (ab-c, c(a^2-1)) \leq 2^{v_2 (a^2-1) }\leq 2a+2 \leq 3a-b $ , so $ c=3 $ and $ a=b+2 \Longrightarrow $ combine $ 3a-b=ca-b \geq 2(bc-a)=6b-2a $ we get $ a=7, b=5 $ . $ \boxed{ \text{ If a is odd. b, c are even. } } \Longrightarrow bc-a=1 \Longrightarrow bc^2-b-c=ca-b $ Since $ c^3-b-c=(1-c^2)(ab-c)+a(bc^2-b-c)+(ca-b) $ , so $ bc^2-b-c=ca-b=\gcd(ab-c, ca-b) \leq \gcd(ab-c, c^3-b-c) $ . $ \text{case 1 :} $ $ c^3-b-c \neq 0 $ From $ b^2c-b-c \leq \gcd(ab-c, c^3-b-c) \Longrightarrow |c^3-b-c| \geq bc^2-b-c $ , so $ b=c \Longrightarrow a=c^2-1 \Longrightarrow ab-c=c(c^2-2) \Longrightarrow c^2-2=2 \Longrightarrow a=3, b=c=2 $ . $ \text{case 2 :} $ $ c^3-b-c = 0 \Longrightarrow b=c^3-c $ From $ bc-a=1 \Longrightarrow a=c^4-c^2-1 $ , so $ ca-b=c^5-2c^3=c^3(c^2-2) \Longrightarrow c^2-2=2 \Longrightarrow c=2 \Longrightarrow a=11, b=6 $ . ____________________________________________________________ From the discussion above $ \Longrightarrow (a,b,c)=(2,2,2), (2,2,3), (2,6,11), (3,5,7) $ and permutations

[Moderator says: do not quote the entire post before yours] a good idea!

TelvCohl wrote: so $ b^2c-b-c=ca-b=\gcd(ab-c, ca-b) \leq \gcd(ab-c, c^3-b-c) $ . I think it should be $bc^2-b-c=ca-b = \gcd (ab-c,ca-b) \le ...$

where can I find the official solutions ?

Why are the inequalities $\gcd(ab-c,ca-b) \leq \gcd (ab-c, c(a^2-1))$ and $2^{v_2 (a^2-1) }\leq 2a+2$ true? Thanks a lot!

This problem is disgusting! One of the worst IMO problems I've ever seen. Despite a good-looking statement, it comes to analyzing a heap of tedious cases. The only trick in the most complicated case ($a>b>c$) is to note that if $ab-c=2^{p+q}$, $ca-b=2^p$, $p,q\in \mathbb Z$, $p \ge 0$, $q \ge 1$, then $$b+c=\frac {2^p(2^q+1)}{a-1},\quad b-c=\frac{2^p(2^q-1)}{a+1}$$and analyze these fractions depending on the largest power of $2$ dividing $a-1$ or $a+1$. This again leads to three cases: $a=2k$, $a=4k+1$ and $a=4k+3$. Each of the fractions breaks into two divisibilities (when $a$ is odd) and after solving them and expressing $a,b,c$ in terms of some other integer parameters, it remains to analyze them based mainly on the inequality $a>b$. What can I say? What is a distinguishing feature of a good programmer (I mean the ability to make a thorough case analysis), is not such for a good mathematician. At least it is a very bad criterion for measuring his abilities on a competition of such rank as IMO. I suggest the IMO Jury to never select such problems in future again.

Bandera wrote: I suggest the IMO Jury to never select such problems in future again. Ummm maybe there's a reason why we are given 4.5 hours per day to solve 3 questions?

Hydrogen-Helium wrote: Ummm maybe there's a reason why we are given 4.5 hours per day to solve 3 questions? I'm pretty sure that there were many other, more beautiful than this one, problems in the shortlist (including number theory) which couldn't be solved by most of contestants in 1.5 hours.

So let's see what happens when two of them are equal, say $a = b$. Then we have $ac-a$ and $a^2-c$ both powers of $2$. Then $a(c-1)$ is a power of $2$ so $a = 2^r$ and $c = 2^s+1$ for some $r, s$. Then that means $2^{2r} - 2^s - 1 = 2^t$ for some $t$. Now, we clearly need to have one of $r, s, t$ be $0$. Clearly $r = 0$ is absurd, so if $s = 0$ we get $4^r = 2^t + 2$ or $2^{2r-1} = 2^{t-1} + 1$. This implies $t = 1$ since the LHS is even, so $r = 1$. Tracing back, we get the solution $(a, b, c) = (2, 2, 3)$. Otherwise, we have $t = 0$ and thus we have $2^{2r-1} = 2^{s-1} + 1$ and so $s = 1$ and so $r = 1$ and so we trace back to get $(a, b, c) = (2, 2, 2)$. Now we can assume $a > b > c$ WLOG, and let $ab - c = 2^z$, $ac - b = 2^y$ and $bc-a = 2^x$, which clearly implies $z > y > x$. We can add the first and second and subtract the first and second to get two equations: $(a-1)(b+c) = 2^y(2^{z-y} + 1)$ and $(a+1)(b-c) = 2^y(2^{z-y} - 1)$. Notice that the gcd between $a-1$ and $a+2$ is at most $2$ which means one of them is not divisible by $4$. Therefore, $2^{y-1}$ divides $b+c$ or $b-c$, so either way, we have $$\frac{ac - b}{2} = 2^{y-1} \le b+c$$and so we get $ac \le 2b + 2c$ or $c \le \frac{2b}{a} + \frac{2c}{a} < 4$. So now either $c = 1, 2, 3$. $c=1$ is bad because then $a-b$ and $b-a$ are both powers of $2$ which makes no sense. If $c = 2$ we have that $ab = 2^z + 2$, $2a - b = 2^y$ and $2b-a = 2^x$. We can use the second and third equations to solve for $a$ and $b$: $$(a, b) = \left( \frac{2^{y+1} + 2^x}{3}, \frac{2^{x+1}+2^y}{3} \right)$$So first of all, we need to make sure that these guys are integers, so we have to have $2^{x-y-1} \equiv -1 \pmod{3}$ which means $x \equiv y \pmod{2}$. Now, we must have $$2^{x+y+1} + 2^{2x} + 2^{2y} + 2^{x+y-1} = 9(2^{z-1} + 1)$$So now notice that the RHS is odd. We therefore must have $x = 0$ (remember that $y > x$). So now we have $$2^{y+1} + 2^{2y} + 2^{y-1} = 9 \cdot 2^{z-1} + 8$$OK so on the LHS the minimal power of $2$ is $2^{y-1}$ so either $y-1 = 3$ or $y-1 = z-1$. We already covered the case of $y = z$ ( because this implies $b = c$), so let's look at $y = 4$. Then $32 + 256 + 8 = 9 \cdot 2^{z-1} + 8$ so $z = 6$. So we have $(x, y, z) = (0, 4, 6)$ and back-tracking yields $(a, b, c) = (11, 6, 2)$. And finally let's look at $c = 3$. This would give us $ab = 2^z + 3$, $3a - b = 2^y$ and $3b - a = 2^x$. We solve with the second and third to get $$a = \frac{3 \cdot 2^y + 2^x}{8} \text{ and } b = \frac{3 \cdot 2^x + 2^y}{8}$$Now multiplying yields $$\frac{10 \cdot 2^{x+y} + 3\cdot 2^{2y} + 3\cdot 2^{2x} }{64} = 2^z + 3$$So then we have $5 \cdot 2^{x+y-5} + 3 \cdot 2^{2y - 6} + 3 \cdot 2^{2x-6} = 2^z + 3$ and so we must have the LHS be odd. Now, if we have exponents on the LHS being negative, that means two of the exponents must be the same, otherwise we cannot form an integer sum. But that is clearly not possible. Thus, since $2x-6$ is the smallest, we know that $2x-6 = 0$ and so $x = 3$. Thus, $$5 \cdot 2^{y-2} + 3 \cdot 2^{2y-6} = 2^z$$Since $y > 3$ we know that $y-2 \le 2y-6$ and so factoring out yields $$2^{y-2} (5 + 3 \cdot 2^{y-4}) = 2^z$$OK so now we see that either $z = y-2$ because $1+3 \cdot 2^{y-4}$ is odd, which will lead to a clear contradiction, or we have $y = 4$ and so then $z = 5$. Tracing back, we have $(x, y, z) = (3, 4, 4)$ and so $(a, b, c) = (7, 5, 3)$. HOORAY! Our solutions are therefore $(2, 2, 2); (2, 2, 3); (3, 5, 7); (2, 6, 11)$

Honestly, I don't know why you guys hate this problem so much. In my opinion, this is my favorite problem I have ever solved. It has an extremely concise problem statement, a great solution set (I just love arbitrary solution sets), it also uses no advanced methods, but uses a lot of different techniques. Maybe not the best in a competition setting (Took me way too long), but an amazing problem for practice. I've always liked case bashing, especially when each of the cases are unique from each other. I also love it when problems have unexpected solutions. What kind of sad problem would this be if you spent hours trying just to prove that the boring $(2,2,2), (3,2,2)$ are the only solutions? Monsters are the fun, surprising parts of math, and keep you on your toes. So anyways:

What I wonder about this problem is why people are assuming that there is not a nice, clean, and elegant solution somewhere out there...

gradysocool wrote: I've always liked case bashing, especially when each of the cases are unique from each other. I also love it when problems have unexpected solutions. What kind of sad problem would this be if you spent hours trying just to prove that the boring $(2,2,2), (3,2,2)$ are the only solutions? Monsters are the fun, surprising parts of math, and keep you on your toes. I guess it's a matter of taste. I personally like pathology and surprising final answers as well, but my complaint is that none of the cases are interesting or even quick. If even one of the cases had some substance to it, I would have liked this problem a lot more. acegikmoqsuwy2000 wrote: What I wonder about this problem is why people are assuming that there is not a nice, clean, and elegant solution somewhere out there... TelvCohl's solution (post #23) is the cleanest I've seen. I have no reason to believe a significantly more elegant solution than that exists, because no one has found one in the 1.5 years since this problem was posed, nor is one included in the official shortlist solutions; nor does it seem likely that any solution can avoid cases and still generate an answer as pathological as $(2,6,11)$.

v_Enhance wrote: gradysocool wrote: I've always liked case bashing, especially when each of the cases are unique from each other. I also love it when problems have unexpected solutions. What kind of sad problem would this be if you spent hours trying just to prove that the boring $(2,2,2), (3,2,2)$ are the only solutions? Monsters are the fun, surprising parts of math, and keep you on your toes. I guess it's a matter of taste. I personally like pathology and surprising final answers as well, but my complaint is that none of the cases are interesting or even quick. If even one of the cases had some substance to it, I would have liked this problem a lot more. Hmmm I actually didn't mind this problem too much as I felt the cases weren't actually too bad. I also liked how you could apply very simple ideas to prove that one of the variables was less than $4$. It was quite a nice number theory/Diophantine problem, which is good as they don't come up very often these days, with a symmetrical and very simple problem statement.

After seeing the number theory problems from shortlist 2015, I like problem N4. Why doesn't N4 get selected? Is it similar to some old problems from olympiad or magazines?

Here's a nice solution that utilises the initial bound throughout, starting off the same as that of post #8. If $2$ of them are equal then we get the solutions $(2,2,2),(2,2,3)$ and permutations. WLOG $1<a<b<c$, we get $2(a+b)\ge ca-b\implies \boxed{3b\ge a(c-2) \cdots (\star)}$, from this $3b\ge a(b-1)\implies a<4$. Suppose $a=2$. If $b,c$ are both even, $v_2(bc-2)=1$, since $b,c>2, bc-2$ is not a power of $2$, contradiction. If $b$ is odd, then $2c-b>1$ is odd and not a power of $2$. Hence $c$ is odd $\implies 2b-c=1$, sub into $(\star)$ we get $3b\ge 2(2b-3)\implies b\le 6$. Checking $(b,c)=(4,7), (6,11)$ we get $(a,b,c)=(2,6,11)$ as a solution. Suppose $a=3$. Sub into $(\star), b\ge c-2$. Since $b,c>3\implies bc-3>1$, so $b,c$ are both odd $\implies b=c-2$. Hence $3b-c=2c-6, 3c-b=2c+2$ are both powers of $2$, $c=7$ is clearly the only solution. Thus we get $(a,b,c)=(3,5,7)$ as another solution and we are done.

Too easy for the IMO... only took me an afternoon. Without further ado, here's the ugliest solution to this problem you'll ever see. Watch and learn:

First, consider the case when at least one of the numbers is divisible by $2$. Also, suppose $v_2(a)\le v_2(b)\le v_2(c)$. Of course, we must have that $v_2(cb)>v_2(a)$, so $bc-a=2^{v_2(a)}$. Also, $v_2(ca)\ge v_2(b)$, with equality if and only if $v_2(a)=0$ and $v_2(b)=v_2(c)$. If equality is not achieved, we also have that $ac-b=2^{v_2(b)}$. Letting the odd parts of $a,b,c$ be $p,q,r$ respectively, we have that $qr|p+1$ from the first equation, and $pr|q+1$. Simple bounding shows that $r<3$. So, $r=1$, and we get $(p,q)=(1,1)$. Now, we know that $a,b,c$ are all powers of $2$, and it is not hard to arrive at the equations: $v_2(a)+v_2(b)=v_2(c)+1$, $v_2(a)+v_2(c)=v_2(b)+1$, $v_2(b)+v_2(c)=v_2(a)+1$, which yields the only triple to be $(2,2,2)$. Now, if we have $v_2(a)=0$ and $v_2(b)=v_2(c)$, we have that $bc=a+1$. Using similar notation as above, we need that $aq-r$ and $ar-q$ are both powers of $2$. First, if $q=r$, then we get that both numbers are divisible by $q$. This means that $q=r=1$, and it follows that $a=2^k+1$. As we also need $a+1$ to be a power of $2$, we must have $a=3$. So, this yields $(2,2,3)$, which works. If $r< q$, then $ar-q|aq-r$, and standard manipulation yields $ar-q|q^2-r^2=(q-r)(q+r)$. As $r$ is an odd number, one of these numbers will be $2\pmod 4$, so $\frac{ar-q}{2}$ divides one of them. In particular, we can get that $ar-q\le 2q+2r$, since we know that $q\neq r$. This gives $r(a-2)\le 3q$. However, we know that $a\ge 4qr-1$, so the RHS is larger than $r(4qr-3)$. Rearranging gives that $3r\ge q(4r^2-3)$, and we see that we will get a contradiction unless $r=1$. In this case, $q\le 3$, so $q=3$. Now, $a-3$ and $3a-1$ are both powers of $2$. As $2(a-3)<3a-1<8(a-3)$ for $a\ge 5$, we must have $4a-12=3a-1$, which yields $a=11$. So, we get the triple $(2,6,11)$, which does work. Finally, if all 3 numbers are odd, then it is clear that none of them can be $1$, or else we get a nonpositive number, and they are all coprime, or else one of the three numbers will have an odd factor greater than $1$. So, we will let $a<b<c$, which means $ab-c<ac-b<bc-a$. We know that $ac-b|bc-a$, so $ac-b|abc-a^2\implies ac-b|b^2-a^2=(b-a)(b+a)$. Like before, $ac-b\le 2(b+a)$, so $2a\ge ac-3b>b(a-3)$. Of course, then, $a=3$. Now, we have $3b-c|3c-b$, so $3b-c|8b$. As $b$ is odd, $3b-c|8$. So, we have $b=\frac{c+2}{3},\frac{c+4}{3}$, or $\frac{c+8}{3}$. Testing them one by one, we get that only $b=\frac{c+8}{3}$ works, which yields $c=7$. So, we get the triple $(3,5,7)$. In conclusion, our 4 solutions are $(2,2,2)$, $(2,2,3)$, $(2,6,11)$, and $(3,5,7)$.

Ok don't mind me asking, but why would someone choose this as an IMO P2... it's like the opposite of, say, IMO 2011 P2 (this comes to mind because of the recent video by 3B1B on this problem)- an elegant statement, but a really, really, case-bashy solution with nothing much to it except for the repeated (and careful) application of $v_2$ and bounding (and a lot of stamina)...(for reference, my solution was rather similar to that of #9's)

ubermensch wrote: Ok don't mind me asking, but why would someone choose this as an IMO P2... it's like the opposite of, say, IMO 2011 P2 (this comes to mind because of the recent video by 3B1B on this problem)- an elegant statement, but a really, really, case-bashy solution with nothing much to it except for the repeated (and careful) application of $v_2$ and bounding (and a lot of stamina)...(for reference, my solution was rather similar to that of #9's) I'd say that this should definitely not appear as P1 since long answers = getting stuck on P1 just writing it up. But, as a P2, it's okay. The difficulty is definitely not P3 level, and one can always skip it for P3. It's more of a question of whether these problems should appear at all...

In retrospect, this was a very poor write-up and hard to understand so I might fix it up sometime soon

The answers are symmetric permutations of: $(2,2,2)$, $(3,2,2)$, $(7,5,3)$, and $(11,6,2)$. It is easy to check that each works, now we show they are the only solutions. First, remark that none of $a,b,c$ is $1$ because if, say, $a=1$, then one of $b-c$ and $c-b$ is nonpositive, absurd. WLOG suppose $a\ge b\ge c>1$. Then, we have \[ab-c\ge ca-b\ge bc-a.\]Note this implies $bc-a\mid ca-b\mid ab-c$. Observe that \begin{align*} bc-a&\mid ca-b\implies bc-a\mid abc-b^2\implies bc-a\mid a^2-b^2\\ bc-a&\mid ab-c\implies bc-a\mid abc-c^2\implies bc-a\mid a^2-c^2\\ ca-b&\mid ab-c\implies ca-b\mid abc-c^2\implies ca-b\mid b^2-c^2. \end{align*}We now observe a few results about parity to split into cases. If $a$ is odd, then either $b$ and $c$ are odd or $bc-a=1$. If $a$ is even, then either $b$ and $c$ are both odd (in which case $bc-a=1$ again) or $bc-a$ is even implying $b,c$ are both even. $a,b,c$ are all odd. Observe that we cannot have $\gcd(b,c)>1$, otherwise $\gcd(b,c)=d$ is an odd positive integer greater than $1$ dividing a power of two. Likewise, $\gcd(a,b)=\gcd(a,c)=1$. Then, note that $\gcd(b+c,b-c)=\gcd(b-c,2c)=2\gcd(b-c,c)=2$, so \[ca-b=\gcd(ca-b,b^2-c^2)\le 2(b+c).\]This rearranges to yield $ca\le 3b+2c$, so \[a\le\frac{3b}{c}+2.\]For $c\ge 5$, observe that \[a\le \frac{3b}{c}+2\le \frac{3b}{5}+2<\frac{3b}{5}+\frac{2b}{5}=b,\]absurd. Hence, we have $c=3$ and $a\le b+2$ implying $a=b+2$ (because $a\ne b$). Then, $ab-c=b(b+2)-3=b^2+2b-3=(b+1)^2-2^2$ is a power of two, so $b-1$ and $b+3$ are powers of two. This extracts the solution $(7,5,3)$. $a,b,c$ are all even. Let $\nu_2(a)=x,\nu_2(b)=y,\nu_2(c)=z$. Since $x,y,z>0$, at most one of $x+y\le z,x+z\le y,y+z\le x$ holds. Then, at least two of the statements $\nu_2(ab-c)=\nu_2(c),\nu_2(ca-b)=\nu_2(b),\nu_2(bc-a)=\nu_2(a)$ hold. Note that $\nu_2(ab-c)=\nu_2(c)$ implies $ab-c\le c\implies ab/2\le c$ because $ab-c$ is a power of $2$, etc. Hence, we must have $ac/2\le b,bc/2\le a$ regardless of which of the statements are true. Then, we have \[a\ge bc/2\ge (ac/2)c/2=ac^2/4\implies 1\ge c^2/4,\]so $c=2$. Then, we have $a\le b$, so $a=b$. Since $ab-2=a^2-2$ is a power of $2$, say $2^k$, we get $a^2=2+2^k$. If $k\ge2$ this is absurd because $2$ is not a QR modulo $4$, and if $k=0$, $3$ is not a perfect square. Hence $k=1$ and $a=2$, so we extract the triple $(2,2,2)$. $bc-a=1$. Rewrite as $a=bc-1$, so we effectively have eliminated the variable $a$; our starting condition is that $(bc-1)b-c=b^2c-b-c$ and $(bc-1)c-b=bc^2-b-c$ are powers of two. We now resolve the case $b=c$. In this case, $bc^2-b-c=b^3-2b=(b^2-2)b$, so $b^2-2$ is a power of two and similarly to before, we get $b=2$ yielding the triple $(3,2,2)$. Then, write \[bc^2-b-c\mid (b^2c-b-c)-(bc^2-b-c) = (b-c)bc.\]Note that $\gcd(bc^2-b-c,bc)\mid b+c$, so \[bc^2-b-c\mid (b-c)(b+c)=b^2-c^2.\]Keeping these observations in mind, we resolve some cases on $b,c$. $b,c$ are both odd. Then, our first observation rewrites as $bc^2-b-c\mid b-c$, which is absurd because $bc^2-b\ge 9b-b>b$. Exactly one of $b,c$ is odd. Then $bc^2-b-c$ must be odd so it is $1$. But similarly $b^2c-b-c=1$, which yields a contradiction because we must have $b>c$. So then $b,c$ are both even. Let $\gcd(b,c)=d$ and note $d$ must be a power of two because $d\mid b^2c-b-c$, so let $d=2^k,a=2^km,b=2^kn$ with $k\ge1$. Then, we have that \[2^{2k}m^2n-m-n,2^{2k}mn^2-m-n\]are powers of two. If only one of $m,n$ was divisible by two, then both expressions would have to be $1$ implying $m=n$, so then $m$ and $n$ are both not divisible by $2$. Write \[2^{2k}mn^2-m-n\mid 2^{2k}(m^2n-mn^2)\implies 2^{2k}mn^2-m-n\mid 2^{2k}(m-n)\]because $2\nmid m,n$. This is absurd for $n\ge3$ because we would get \[2^{2k}(m-n)-2^{2k}mn^2+m+n\le 2^{2k}m-2^{2k}n-9\cdot2^{2k}m+m+n<0.\]Hence, we have $n=1$. Then, we have \[2^{2k}m-m-1\mid 2^{2k}m-2^{2k}\implies 2^{2k}m-m-1\mid m+1-2^{2k}\implies 2^{2k}m-m-1\mid 2^{2k}-m-1.\]Since $2^{2k}m>2^{2k}$, either $2^{2k}<m+1$ or $2^{2k}=m+1$. The former case is not a possibility because then $2^{2k}m-m-1\ge 3m-1$ but $|2^{2k}-m-1|<|m-3|$. Then, we have that \[2^{2k}m-m-1=(m+1)m-m-1=m^2-1=(m-1)(m+1)\]is a power of two, so $m=3$. This extracts $k=1$, so we obtain the solution $(11,6,2)$ to finish.

IMO 2015 P2 wrote: Find all positive integers $(a,b,c)$ such that $$ab-c,\quad bc-a,\quad ca-b$$are all powers of $2$. Proposed by Serbia Let $ab-c=2^x, bc-a=2^y,ca-b=2^z$ with $x,y,z$ being non-negative integers. We have two cases to consider. Case 1: At least two of $a,b,c$ are equal. WLOG assume $b=c$. Then, $ab-b=2^x$ and $b^2-a=2^y$. If $x=0$, then $b(a-1)=1$, therefore $a=2,b=1$, so $2^y=-1$, which is a contradiction. If $y=0$, then $a=b^2-1$. Therefore, $b(b^2-2)=2^x$. If $b=1$ we have an obvious contradiction. Hence $b \geq 2$, implying that $b$ is even. Thus $b^2-2 \equiv 2 \pmod 4$, hence $b^2-2=2$. Therefore $a=3,b=2$, hence the solution $(3,2,2)$. Suppose now that $x,y \geq 1$. Then, $a-1$ divides $2^x$ so $a=2$ or $a$ is odd. If $a=2$ we get $b=2^x$ and $b^2=2^y+2$, hence $x,y$ can't be both $\geq 2$ due to $\pmod 4$ reasons. Therefore some easy case bash implies $(a,b,c)=2$. If now $a$ is odd, then $b^2=a+2^y$ so $b$ is odd as well. Since $b \mid 2^x$, this implies $b=1$, which is a clear contradiction since $b^2=a+2^y >1$. Case 2: $a \neq b \neq c \neq a$. WLOG suppose $a>b>c$, which in turn implies $x>z>y$. We go into subcases. Subcase 2.1: $y=0$. Then, $a=bc-1$, hence $$(bc-1)b-c=2^x, (bc-1)c-b=2^z \,\, (\bullet)$$with $x>z \geq 1$, If at least one of $b,c$ is odd we easily get a contradiction. Hence $b,c$ are both even. Let $v_2(b)=k, v_2(c)=\ell$ with $k,\ell \geq 1$. Subtract and sum the two equations in $(\bullet)$ to obtain $$bc(b-c)=2^z(2^{x-z}-1) \,\, (*)$$and $$(bc-2)(b+c)=2^z(2^{x-z}+1) \,\, (**)$$Suppose firstly that $k \neq \ell$. Then, note that $$v_2(b+c)=v_2(b-c)=\min(k,\ell)$$hence using $(*)$ and $(**)$ we obtain $$z=k+\ell+\min(k,\ell)$$and $$z=\min(k,\ell)+1$$Therefore $k+\ell=1$, which is a contradiction. Conseqently, $k$ and $\ell$ must be equal. Take $b=P \cdot 2^k$ and $c=Q \cdot 2^k$ with $P,Q$ being odd. Substituting the above to $(*)$ and $(**)$ we obtain $$2^{3k}PQ(P-Q)=2^z(2^{x-z}-1)$$and $$2^k(2^{2k}PQ-2)(P+Q)=2^z(2^{x-z}+1)$$Hence we obtain $$z=3k+v_2(P-Q)=k+1+v_2(P+Q) \,\, (***)$$ In addition, $$2^z=c^2b-c-b=2^k(2^{2k}Q^2P-P-Q)$$If $v_2(P+Q) \geq 2k+1$, then using the latter we obtain $z=3k$, which is a contradiction since $$3k=z=k+1+v_2(P+Q) \geq 3k+2$$Therefore, $v_2(P+Q) \leq 2k$, implying $$3k+1 \leq 3k+v_2(P-Q)=k+1+v_2(P+Q) \leq 3k+1$$hence equality must hold. Therefore $z=3k+1, v_2(P-Q)=1,v_2(P+Q)=2k+1$. Hence, $(bc-1)c-b=2^z$ rewrites as $$P+Q=2^{2k}(Q^2P-2)$$Consequently, we have $$P+Q=2^{2k}(Q^2P-2) \geq 4Q^2P-8 \Rightarrow Q+8 \geq (4Q^2-1)P \geq 4Q^2-1 \Rightarrow 4Q^2 \leq Q+9 \Rightarrow Q=1$$Therefore, $$P+1=2^{2k}(P-2) \Rightarrow (P-2) \mid (P+1) \Rightarrow P \in \{3, 5 \}$$Considering both cases we obtain the solution $(a,b,c)=(11,6,2)$. Subcase 2.2: $y \geq 1$. We prove the following Claim. Claim: $a,b,c$ are all odd. Proof: Suppose otherwise. It is easy to see that if one of them is even, then all are even as well. Note that $$2^x+2^z=ab-c+ac-b=(a-1)(b+c)$$and similarly $$(a+1)(b-c)=2^x-2^z$$. Comparing the $v_2$'s we obtain $v_2(b+c)=v_2(b-c)=z$. Therefore, $$2^z \mid (b+c+b-c)=2b \Rightarrow 2^{z-1} \mid b$$and similarly $2^{z-1} \mid c$. Hence $b,c \geq 2^{z-1}$. Therefore, we have the following chain of inequalities: $$2^z=ca-b >cb-b=b(c-1) \geq 2^{z-1}(2^{z-1}-1) \Rightarrow 2^{z-1} <3,$$hence $z \leq 2$. Since $2 \geq z \geq y+1 \geq 2$, we obtain $y=1,z=2$. As a result, $$(c+1)(a-b)=(ca-b)-(bc-a)=2^z-2^y=2 \Rightarrow c=1, a-b=1$$Hence $$2^y=bc-a=b-a=-(a-b)=-(ca-b)=-2^z,$$absurd. Hence we are done $\blacksquare$ To the problem, $a,b,c$ are all odd. To sum up, we have so far obtained that $a>b>c$, $a,b,c$ are odd and $x>z>y \geq 1$. Note that, $$c^2+2^x \cdot c=c(c+2^x)=abc=a^2+2^y \cdot a=b^2+2^z \cdot b$$hence $$a^2-c^2=2^x \cdot c-2^y \cdot a \equiv 0 \pmod {2^y}$$Therefore $2^y \mid (a^2-c^2)$. In addition, $2^y=bc-a \mid (b^2c^2-a^2)$. Since $a^2 \equiv c^2 \pmod {2^y}$, we conlude that $2^y \mid (b^2c^2-c^2) \Rightarrow 2^y \mid (b^2-1)$ since $c$ is odd. Therefore, $2^y \mid (b-1)(b+1)$. Note that $\gcd(b-1,b+1)=2,$ therefore either $v_2(b-1)$ or $v_2(b+1)$ is $\geq y-1$. In any case we may conclude that $b \geq 2^{y-1}-1$. In a similar manner, $a \geq 2^{z-1}-1$ and $c \geq 2^{y-1}-1$. Thus, $$2^z-2=ca-(b+2) \geq ca-a=a(c-1) \geq 2a \geq 2^z-2,$$therefore we have equality everywhere. Note that the first inequality holds since $a>b$ and they are both odd, while for the second one, if $c=1$ then $$2^y=b-a=-(a-b)=-2^z,$$a contradiction. Since $c$ is odd as well, $c \geq 3$. Thereofre, $a=b+2,c=3$ and $a=2^{z-1}-1$. Substituting these to $2^y=bc-a$ we obtain $2^y=2^z-8$ which due to $\pmod 16$ has no solutions if $y \geq 4$. Check some easy cases now to obtain $y=3$ and $z=4$, that is $(a,b,c)=(7,5,3)$. To sum up, the solutions are $(3,5,7), (2,2,2), (2,2,3), (2,6,11)$ up to permutations.

Let $ab-c=2^x$ and $bc-a=2^y$ and $ca-b=2^z$. Case 1: a,b,c not all odd. Let $a=2^{a_1}x_1$ and $b=2^{b_1}y_1$ and $c=2^{c_1}z_1$, where $x_1,y_1,z_1$ are odd. WLOG assume $a_1\ge b_1\ge c_1$, else permute the variables. So $a_1\ge 1$. We have $2^x=ab-c=2^{a_1+b_1}x_1y_1-2^{c_1}z_1$, so $2^{c_1}(2^{a_1+b_1-c_1}x_1y_1-z_1)=2^x$. Comparing $\nu_2$ of both sides, we get $c_1\le x$. So dividing by $2^{c_1}$ on both sides, \[ 2^{a_1+b_1-c_1}x_1y_1-z_1=2^{x-c_1}.\]Note $a_1+b_1>c_1$ since $a_1\ge 1$. So the LHS is even-odd=odd, so we must have $x=c_1$. Also, now, $z_1=2^{a_1+b_1-c_1}x_1y_1-1 \ge 2x_1y_1-1$. We have $2^z=ac-b=2^{a_1+c_1}x_1z_1-2^{b_1}y_1=2^{b_1}(2^{a_1+c_1-b_1}x_1z_1-y_1)$, after noting the fact that $a_1+c_1\ge b_1$, with equality if and only if $c_1=0$. Comparing $\nu_2$ of both sides, we conclude $z\ge b_1$, so dividing by $2^{b_1}$ on both sides, \[2^{z-b_1}=2^{a_1+c_1-b_1}x_1z_1-y_1.\]Now: Subcase 1: $a_1+c_1>b_1$. Then the RHS is even-odd=odd, hence the LHS must be 1. So $z=b_1$. So $y_1=2^{a_1+c_1-b_1}x_1z_1-1 \ge 2x_1z_1-1$. Recall $z_1\ge 2x_1y_1-1$, so now $y_1\ge 2x_1(2x_1y_1-1)-1=4x_1^2y_1-2x_1-1\ge (4x_1^2-2x_1-1)y_1$, so $4x_1^2-2x_1\le 2$, which is impossible unless $x_1=1$. Equality is achieved if $x_1=1$ in all bounds above, so we can easily find $x_1=y_1=z_1=1$. So $a,b,c$ are all powers of 2, so $a_1+b_1-c_1=1$ and cyclic variants by the original equations, so $(a,b,c)=(2,2,2)$. Subcase 2: $a_1+c_1=b_1$. This means $c_1=0$ since $a_1\ge b_1\ge c_1$. So $a_1=b_1$. Using the two inline equations above, we have $z_1=2^{2a_1}x_1y_1-1$ and $y_1=x_1z_1-2^{z-a_1}$. So \[ z_1 = 2^{2a_1}x_1(x_1z_1-2^{z-a_1})-1 \implies z_1=\frac{2^{z+a_1}x_1+1}{2^{2a_1}x_1^2-1}.\]So \[ y_1 = x_1z_1-2^{z-a_1}=x_1\cdot \frac{2^{z+a_1}x_1+1}{2^{2a_1}x_1^2-1} - 2^{z-a_1} = \frac{2^{z+a_1}x_1+1}{2^{2a_1}x_1 - \frac{1}{x_1}}-2^{z-a_1}.\]The above is not an integer unless $x_1=1$! So $x_1=1$. Hence $z_1 = \tfrac{2^{z+a_1}+1}{2^{2a_1}-1}$. So $2^{z+a_1}\equiv -1 \pmod{2^{2a_1}-1}$. In this modulus, $2^{2a_1} \equiv 1$, so $2^{z+a_1}\equiv 2^\ell$ for some $\ell \le 2a_1-1$. So $2^\ell \equiv 2^{2a_1}-2$. But both of these are less than the modulus, so $2^\ell = 2^{2a_1}-2$. So $\ell=1$ and $a_1=1$. So now $a=2$, $c=z_1=\tfrac{2^{z+1}+1}{3}$ and $b=2y_1=\tfrac{2^{z}+2}{3}$. Now $bc-a\approx 2^{2z+1}/9$, which is not a power of 2 for large enough $z$, and we can eliminate $z\ge 5$ with some bounding. Now only $z=2$ and $z=4$ work, which correspond to $(a,b,c)=(2,2,3), (2,6,11)$. Case 2: a,b,c all odd. WLOG $a\le b\le c$, else permute the variables. Then we can easily show $ab-c\le ac-b\le bc-a$, so since powers of 2, the first 2 divide subsequent ones. We claim $\gcd(a,b)=1$ and cyclic variants. Indeed, suppose $k\mid a,b$, then $k\mid ac-b$, a power of 2, so $k$ is a power of 2, but if $k\not = 1$ then it is even, contradiction. Now \begin{align*} ac-b&\mid bc-a \implies ac-b\mid c(ac-b)+(ac-b)=b(c^2-1). \\ ac-b&\mid bc-a \implies ac-b\mid (bc-a)+c(ac-b)=a(c^2-1). \end{align*}Since $\gcd(a,b)=1$, combining the above implies either \[ ac-b\mid c^2-1=(c-1)(c+1) \implies \frac{ac-b}{4} \mid \frac{c-1}{2} \cdot \frac{c+1}{2}, \]or $ac-b=2$ (since it's a power of 2 and is even). Subcase 1: $ac-b\ge 4$. Then using the above divisibility, since $\gcd(\tfrac{c-1}{2},\tfrac{c+1}{2})=1$, in fact $\tfrac{ac-b}{4}$ divides one of the two, since it's a power of 2. This is the key number theoretic property that we are using -- it's a power of 2 -- rather than just divisibility and size. So \[\frac{ac-b}{4}\le \frac{c+1}{2} \implies c\le \frac{b+2}{a-2}. \]Also, $ac-b\mid bc-a\implies ac-b\mid a(bc-a)-b(ac-b)=b^2-a^2$, so $b^2-a^2 \ge ac-b \ge a(a(b-a)+1)-b=(a^2-1)(b-a)$, so $b+a\ge a^2-1$, so \[ b\ge a^2-a-1.\]We can prove similar to the above that $ab-c\mid a^2-1$, so $ab-c\le a^2-1$, so $c\ge a(b-a)+1$. Now, combining everything, \begin{align*} &\qquad b+2 \ge c(a-2) \ge (ab-a^2+1)(a-2) \\ &\implies (a^2-2a-1)b \le a^2(a-2)-(a-2)+2 = a(a^2-2a-1)+4 \\ &\implies b \le a + \frac{4}{a^2-2a-1}. \end{align*}If $a\ge 4$, then the above implies $b=a$. In this case, $c\le \tfrac{b+2}{a-2}=\tfrac{a+2}{a-2}=1+\tfrac{4}{a-2}$. Now we can manually check cases, and we get that no $(a,b,c)$ pop out. So $a\le 3$. If $a=3$, then $b\le 3+2=5$, and manually checking gives that only $(a,b,c)=(3,5,7)$ pops out. If $a=1$, we get nothing. Subcase 2: $ac-b=2$. Then $ab-c=2$ too. So $\tfrac{b+2}{c}=\tfrac{c+2}{b}$, so $b^2+2b=c^2+2c$, so $b=c$. Now we can easily find that nothing works. So the answer is $(a,b,c)=(2,2,2), (2,2,3), (2,6,11), (3,5,7)$, and all permutations.

We claim the solutions are $(2, 2, 2), (2, 2, 3), (2, 6, 11), (3, 5, 7)$ and their permutations. These can be checked to work. Case 1: WLOG $a = b$. $a(c-1) = ac - a$ power of 2 means $a$ and $c-1$ are. Hence Let $a = b = 2^n$, $c = 2^m+1$. $a^2 - c = 2^k \Rightarrow 2^{2n} - 2^m - 1 = 2^k$. If $m, n > 0$, we have $LHS$ odd so $2^k = 1$. Hence $2^{2n} - 2^m = 2 \Rightarrow 2^{2n-1} - 2^{m-1} = 1$ and $m = 1, n = 1$. This gives us $(2, 2, 3)$. Else $m = 0$ so $2^{2n}-2=2^k$ means $n = 1$, giving us $(2, 2, 2)$, or $n = 0$ means $-2^m = 2^k$ is a contradiction. Case 2: Now assume $a < b < c$. Letting $ab - c = 2^\lambda$, $ca - b = 2^m$, $bc - a = 2^n$, we get $\lambda < m < n$. If $a = 1$, $b - c = 2^\lambda$ is negative, contradiction. Hence $a > 1$. \begin{align*} 2^m = ca - b > ca - c = c(a-1) > c. \tag{1} \end{align*}$ab - c | ca - b | bc - a$ means $ca - b | abc - b^2$ and $ca - b | abc - a^2$, so \begin{align*} ca - b | b^2 - a^2 \Rightarrow v_2(b-a) + v_2(b+a) \ge m. \tag{2} \end{align*} Case 2.1: $c$ even. Assume for contradiction $v_2(a) = v_2(b) = r$. Then $v_2(ca - b) = m$, but $v_2(ca) > v_2(a) = r = v_2(b) \Rightarrow v_2(ca - b) = v_2(b) = r$. Hence $v_2(a) = r = m$, and $a \ge 2^m = ca - b$. Thus $a + b \ge ca > ba \ge 2b$ is a contradiction. By (2) we get $2 \min(v_2(a), v_2(b)) \ge m$ so $a \ge 2^{m/2}$ and $ab \ge 2^{m+1}$ (since $v_2(a) \neq v_2(b)$). $2^\lambda = ab - c \ge 2^{m+1} - c \Rightarrow c \ge 2^{m+1} - 2^\lambda$. Now $2^m = ca - b \ge (2^{m+1} - 2^\lambda)(2^{m/2}) - b \ge (2^{m+1} - 2^{m-1})(2^{m/2}) - b \Rightarrow b > 2^{3m/2} - 2^m \ge 2^m$, as $m \ge 1$. But $c > b > 2^m$ contradicts (1). Case 2.2: $c$ odd. Case 2.2.1: $v_2(a) \neq v_2(b)$. Same as above. Case 2.2.2: $v_2(a) = v_2(b) = r \ge 1$. $ab - c$ odd so $2^\lambda = 1$, and $c = ab - 1$. Thus after substitution we get \begin{align*} a^2b &= 2^m + a + b \\ ab^2 &= 2^n + a + b \\ \Rightarrow ab(b - a) &= 2^m(2^{n-m} - 1). \end{align*} $n > m$ means $v_2(a) + v_2(b) + v_2(b - a) = m$. $v_2(b-a) \ge r \Rightarrow n > m \ge 3r$. Thus by considering $v_2(ab^2) = 3r = v_2(2^n + (a+b))$, we need $v_2(a+b) = 3r$. Hence $v_2(a-b) = v_2((a+b)-2b)$, and $3r > r+1$ means $v_2(a+b) > v_2(2b)$ so $v_2(a-b) = v_2(2b) = r+1$. This gives $m = 3r+1$. Now $a^2b - a - b = 2^m \Rightarrow 2^{3r+1} > a^2b - 2b = (a^2-2)b \ge 2b$ so $b < 2^{3r}$. But $v_2(a+b) = 3r \Rightarrow a+b \ge 2^{3r}$, so $2^{3r+1} = 2b > a + b \ge 2^{3r}$. For $v_2(a+b) = 3r$ to hold we need $a + b = 2^{3r}$, so $a^2b = 2^{3r+1} + 2^{3r} = 3\cdot2^{3r}$. If $3 | a$ then $9 | 3 \cdot 2^{3r}$ is absurd so $3 | b$, and $a = 2^r$, $b = 3 \cdot 2^r$. $a + b = 4 \cdot 2^r = 2^{r+2}$, so $3r = r + 2$ means $r = 1$. This gives $(2, 6, 11)$. Case 2.2.3: $a, b, c$ odd. (1) can now be strengthened as $a \ge 3$ means $2^m > 2c$ so $c < 2^{m-1}$. Now note that one of $b-a$, $b+a$ is $2 \pmod{4}$, so either $v_2(b+a) \ge m-1 \Rightarrow b + a \ge 2^{m-1} \Rightarrow b > 2^{m-2}$ or $v_2(b-a) \ge m-1 \Rightarrow b > 2^{m-1}$. Either case, we get $b > 2^{m-2}$. Hence, $2^{m-1} > c > b > 2^{m-2}$. $2^{m-1} > c = ab - 2^\lambda > 3 \cdot 2^{m-2} - 2^\lambda \Rightarrow 2^\lambda > 2^{m-2}$. Hence $\lambda = m - 1$. Now we have $c = ab - 2^{m-1}$. Assume for contradiction $a \ge 5$. Then $c > 5\cdot2^{m-2} - 2^{m-1} = 3 \cdot 2^{m-2} > 2^{m-1} > c$, contradiction. Thus $a = 3$, and $c = 3b - 2^{m-1}$. Substituting into $ca - b = 2^m$ we get $9b - 3 \cdot 2^{m-1} - b = 2^m \Rightarrow 8b = 5 \cdot 2^{m-1}$, so $b = 5 \cdot 2^{m-4}$. $b$ odd means $b = 5$, and we get $(3, 5, 7)$ as our final solution.

It is easy to see if two numbers are equal, the only solutions are $(2, 2, 2), (2, 2, 3)$ and perms. We split the remaining cases by parity. WLOG $a \leq b \leq c$ and the powers of $2$ in increasing order be $2^x, 2^y, 2^z$. Notice $a = 1$ fails immediately. Any perms are implied. If all even, then we lose immediately by $v_2(bc-a) = v_2(a) \Longrightarrow bc-a \geq a$. If one even, two odd, then two powers of two are odd, so one pair is equal giving contradiction. If all odd, then $(c-1)(a+b) = 2^z-2^y, (c+1)(b-a) = 2^z+2^y$ so since either $v_2(a+b)$ or $v_2(b-a)$ is $1$, $c \equiv \pm 1 \pmod 2^{y-1}$. However $ac-b = 2^y$ so $a \leq 3$ and $c = 2^{y-1}-1$ and $b = 2^{y-1}-3$ so by check, the only solution is $(3, 5, 7)$.

Note that if $(a,b,c)$ works then any permutation works. Suppose two of the numbers are the same, say $a=b$ then $a^2-c$ and $ac-a$ are both powers of two. Since $ac-a=a(c-1)$, we can say $a=2^x$ and $c=2^y+1$. We get $2^{2x}-2^y-1$ is a power of two. However, unless $x$ or $y$ is zero, this is odd, which means the only possible value is $1$. In this case, $2^{2x}-2^y=2$ which implies $x=1$, $y=1$. If $x$ is zero, obviously $2^{2x}-2^y-1$ is negative and thus not a power of two. If $y$ is zero, then we have $2^{2x}-2$ is a power of two, implying $x=1$ again. Thus, the solutions are $(a,b,c)=(2,2,2)$, $(2,2,3)$ and permutations. Assume $a$, $b$, $c$ are all distinct then let $a<b<c$. If $a=1$ then $b-c$ and $c-b$ are powers of two. If $a=2$ then consider $\nu_2(bc)$. Clearly, if $\nu_2(bc)\ge 2$ then $\nu_2(bc-2)=1$ so $bc-2=2$, which implies $bc=4$, impossible. If $bc$ is odd then $bc-2$ is odd and thus is equal to $1$, so $bc=3$, also impossible. One is odd and the other is even. If $b$ is odd then $2c-b > b > 2$ is also odd, impossible to be a power of two. If $c$ is odd and $b$ is even then $2b-c$ is odd, so $2b-c=1$. We get that $3b-2$ is a power of two. Let $b=\tfrac{2^k+2}{3}$ and using $b(2b-1)-2$ is a power of two, we get the following is a power of two: \[\frac{(2^k+2)(2^{k+1}+1)-18}{9}=2\left(\frac{2^{2k}+2^{k+1}+2^{k-1}-8}{9}\right)\]If $k\ge 5$ then $\nu_2(2^{2k}+2^{k+1}+2^{k-1}-8)=3$ which leaves the numerator too big and thus creating odd primes. If $k=4$ then $b=6$, $c=11$ which works. If $k=3$ then $b$ is noninteger. If $k\le 2$ then $b\le a$, impossible. Thus, the only other solution with $a=2$ is $(2,6,11)$ and permutations. Suppose $a=3$. We have $3c-b < bc-3$ and both are powers of two, so $3c-b\mid bc-3-3c+b=(b-3)(c+1)$ and $3c-b\mid bc-3+3c-b=(b+3)(c-1)$. Either $c-1$ or $c+1$ is not divisible by $4$, so either $2b+6$ or $2b-6$ is divisible by $3c-b$. Either way, $3c-b\le 2b+6$ so $c\le b+2$. Clearly, both $b$ and $c$ must be odd, otherwise $bc-3$ is odd which requires $bc=4$, absurd. Thus, $c=b+2$. We need $bc-3=b^2+2b-3=(b+3)(b-1)$ be a power of two, so we must have $b-1=4$ and $b+1=8$, so $b=5$, $c=7$. This gives our final solution, $(3,5,7)$ are permutations. Suppose $a\ge 4$. As before, we will have $ac-b\le 2b+2a$ so $4c-8\le a(c-2)\le 3b\le 3c-3$. Thus, $c\le 5$ which is a contradiction. We are done.

The only solutions are $(a,b,c) = (2,2,2), (3,2,2), (7,5,3), (11,6,2)$ and permutations. These work. Now we prove they are the only ones. Claim: If there exist two equal numbers in $(a,b,c)$, then we either have $a = b = c = 2$ or $(a,b,c) = (3,2,2)$ or permutations. Proof: WLOG that $b = c$. Then $b(a-1)$ and $b^2 - a$ are powers of $2$. If $b = 1$, then $b^2 - a$ cannot be a power of $2$. Therefore, $b$ must be even. If $a = 2$, then $b^2 - 2$ is a power of $2$, so $b = 2$. If $a > 2$, then $a = 2^k + 1$ for some $k>0$, meaning $b^2 - (2^k + 1)$ must be a power of $2$. But since $b$ is even and $2^k + 1$ is odd, $b^2 - (2^k + 1) = 1$, so $b^2 = 2^k + 2$. Since $b$ is a power of $2$, $b^2$ and $2^k$ are powers of $2$ with difference two, so one of them must equal $2$. Therefore $2^k = 2$ and $b^2 = 4$, which means $a = 3$ and $b = 2$. $\square$ Additionally, if one of the variables is $1$ (say $c$), then $b - a$ and $a - b$ are powers of $2$, impossible. WLOG that $a > b > c \ge 2$ and let $2^x = ab - c, 2^y = ac - b, 2^z = bc - a$. We have $x > y > z$ and $2^x - 2^y = (ab - c) - (ac - b) = (a+1) (b - c)$. Now, we can do the following: $2^x = ab - c = a(ac - 2^y) - c$. Taking modulo $2^y$, we get that $2^y \mid c(a^2 - 1)$. We also have $2^y = ac - b = a(ab - 2^x) - b$. Now taking modulo $2^x$, we see that $a^2 b - b = b(a^2 - 1)$ is $2^y$ modulo $2^x$, so $\nu_2(b(a^2 - 1)) = y$. Thus we have that \[ \nu_2(c(a-1)(a+1)) \ge y, \nu_2(b(a-1)(a+1)) = y \ \ \ \ \ \ \ \ \ \ \ (1)\] Case 1: $a,b,c$ are all odd. Then, we have that $\nu_2((a-1)(a+1)) = y$ since $\nu_2(b) = 0$. Since $a$ is odd, exactly one of $a-1$ and $a + 1$ has a $\nu_2$ of $1$. The other one must have a $\nu_2$ of $y-1$. This implies that $a \ge 2^{y-1} - 1$. Hence, $a \ge \frac{ac - b}{2} -1$, which means $2a \ge ac - b - 2$, so $b+2 \ge a(c-2)$. Since $a,b$ are odd, we have $a \ge b+2$, so $a \ge a(c-2)$, so $c = 2$ or $c = 3$. Since $c$ is odd, we must have $c = 3$ and $b = a - 2$. Now, $bc - a = 3(a-2) - a = 2a - 6$ and $ac - b = 2a + 2$ are powers of $2$ with a difference of $8$, meaning $2a - 6 = 8$, so $a = 7$ and $b = 5$, giving $(a,b,c) = (7,5,3)$. Case 2: Two of $a,b,c$ are odd, and the third is even. This implies that two of $ab - c , bc - a, ca - b$ are odd, so two of them must equal $1$, which is impossible since $2^x > 2^y > 2^z$. Case 3: $a,b,c$ are all even. Then $(1)$ gives that $\nu_2(c) \ge y$ and $\nu_2(b) = y$. This means that $b \ge 2^y = ac - b$, so $2b \ge ac$. But $ac > bc$, meaning $2b > bc \implies 2 > c$, absurd. Case 4: One of $a,b,c$ is odd, and the other two are even. Claim: $a$ must be odd. Proof: Suppose $a$ was even. If $c$ was odd, then $ab - c = 1$, so $ab = c + 1$, which is clearly impossible since both $a$ and $b$ are at least $c + 1$. If $b$ was odd, then $ac - b = 1$, so $ac = b + 1$, which is impossible since $ac \ge 2a > b + 1$. Therefore $a$ must be odd. $\square$ This implies that $bc - a = 1$, so $a + 1 = bc$. Since $b$ and $c$ are even, $a$ must be $3\pmod 4$. Now we prove a claim relating to the $\nu_2$s of the numbers. Claim: $\nu_2(b) = \nu_2(c)$ Proof: Suppose this was not the case. By $(1)$, we get that $\nu_2(c) \ge \nu_2(b)$, so $\nu_2(c) > \nu_2(b)$. We have $\nu_2(a+1) = \nu_2(b) + \nu_2(c)$ and $\nu_2(b-c) = \nu_2(b)$. Since $2^x - 2^y = (a+1) (b-c)$, we have that $\nu_2((a+1)(b+c)) \ge y$, so $2\nu_2(b) + \nu_2(c) \ge y$. However, from $(1)$ again, we see that since $\nu_2(a-1) = 1$, $y =\nu_2(b(a-1)(a+1)) = 2\nu_2(b) + \nu_2(c) + 1$, which means $2 \nu_2(b) + \nu_2(c) < y$, contradiction. $\square$ Notice that $y = \nu_2(b(a-1)(a+1)) = 2\nu_2(b) + \nu_2(c) + 1 = 3\nu_2(b) + 1$. Let $m = b + c$. We have $\nu_2(m) > \nu_2(b)$. Now since $a = bc - 1$, we have $2^x = ab - c = b^2 c - m$ and $2^y = ac - b = b c^2 - m$. If $\nu_2(m) < 3\nu_2(b)$, then we would have $\nu_2(b^2 c - m) = \nu_2( bc^2 - m) = \nu_2(m)$, which is a contradiction since $x\ne y$. Hence $\nu_2(m) \ge 3\nu_2(b) = y - 1$. Thus, $m \ge 2^{y-1} = \frac{b c^2 - m}{2}$. This implies that \[ 2m \ge bc^2 - m\implies 3m \ge bc^2 \]If $c \ge 4$, then $3m \ge 16b$, but $3m = 3(b+c) \le 6b$, absurd. Since $c$ is even, we must have $c = 2$, so $3(b+2) \ge 4b\implies b \le 6$. Now, $a + 1 = 2b$, so $a = 2b - 1$. We have $2a - b = 3b - 2$ must be a power of $2$. Since $b \ge 3$, we can check that $3b - 2$ is not a power of $2$ when $b\in \{3,4,5\}$, so $b = 6$ must hold. Hence $a = 2b - 1 = 11$, giving $(11,6,2)$.

Due to symmetry, we can also consider $a\geq b\geq c$ with $ab-c=2^k, bc-a=2^l, ca-b=2^m$. We will initially assume that $k,l,m>0$ with $k\geq m\geq l$. Subtracting the third from the first, we get: $ab-c-ca+b=2^m(2^{k-m}-1)\Leftrightarrow (a+1)(b-c)=2^m(2^{k-m}-1) \ (1)$. We will distinguish $2$ cases: $\bullet$ Then, if $b=c$, it will be true $ab-b=2^k\Leftrightarrow b(a-1)=2^k$. If $a-1$ is odd, it must be $1$, resulting in $a=2$ and $b=c=2^k$. With substitution in the second equation, we get $b^2=2^l+2$, which leads to $2^{2k}=2^l+2$, $l\leq k$. So, $2^{2l}-2^l-2\leq 0\Leftrightarrow (2^l-2)(2^l+1)\leq 0$. Therefore, $2^l=1$ (rejected) or $2^l=2$. In this case, we get the solution $\boxed{(a,b,c)=(2,2,2)}$. Then, $b\neq c$, since $a+1$ is superfluous, it holds $2^m\mid b-c$. Therefore, $2^m\leq b-c\Leftrightarrow ca-b\leq b-c\Leftrightarrow c(a+1)\leq 2b$ (due to the chosen layout), which is possible only if $c=1$, leading to no solutions. $\bullet$ Suppose $a$ is superfluous; the other $a+1,a-1$ can be divisible at the same time by $4$. Suppose $4\nmid a+1$. Then, according to $(1)$, $2^{m-1}\mid b-c$. If $b=c$, then $ab-b=2^k\Leftrightarrow b(a-1)=2^k$. For $b=1$, there are no solutions. For $b=2^n$, $b^{2}-2=2^{2n}-2$. But $b^2-2=2^p$, so $2^{2n}-2^p=2$, yielding $2$ forces that make a difference. So, $b=2$ leads to the solution $\boxed{(a,b,c)=(3,2,2)}$, but in the first instance, it is rejected since we assumed that $4\nmid a+1$. Therefore, $b\neq c$, and $2^{m-1}\mid b-c$. If $b\neq c$, then $2^{m-1}\mid b-c$, $b-c\geq \frac{ca-b}{2}\Leftrightarrow 3b\geq c(a+2)$ (due to the provision we have chosen), so $c\leq 2$. If $c=1$, there are no solutions; if $c=2$, the second equation gives $2b-a=2^l$. i.e., $a$ would be even, which is out of place. Let us now assume that $4\nmid a-1$. It is superfluous $bc=2^l+a$, and therefore superfluous $b,c$. Adding the first to the third ratio, we get $(a-1)(b+c)=2^m(2^{k-m}+1) \ (2)$. It will necessarily be the case $2^{m-1}\mid b+c$. $b+c\geq \frac{ca-b}{2}\Leftrightarrow c(a-2)\leq 3b$ due to the provision we have chosen, it follows that $c\leq 5$. $\bullet$ If $c\geq 5$, then $5(a-2)\leq c(a-2)\leq 3b\Leftrightarrow 5a-3b\leq 10\overset{a\geq b}\Leftrightarrow 2b\leq 5a-3b\leq 10\Leftrightarrow b\leq 5$. So, it would be true $b<c$, which is out of place. Therefore, $c\leq 5$. $\bullet$ If $c=5$, then identical to before, we end up with the solution $a=b=c=5$, which is not acceptable. $\bullet$ If $c=1$, then obviously, there are no solutions. $\bullet$ If $c=3$, then $ab=2^k+3$ and $2b-a=1$, $2a-b=2^m$. If both are redundant, then from the third relationship, we get $2^m$ redundant, so $2a-b=1$. But $a=b=1$ is also true, but we do not get solutions from it. If any of them is even, then $ab$ is even, therefore $2^k+3$ is even, therefore $k=0$. But then it necessarily applies (because of the provision) and $m=0$; therefore, we get again $a=b=1$, which is out of place. . $\bullet$ If $c=4$, then $ab=2(2^{k-1}+2)$ and $4b-a=1$, $4a-b=2^m$. Because $a=4b-1$, the second relationship is written $16b-4-b=2^m\Leftrightarrow 15b=2^m+4$. Obviously, $m\neq 0$, we take it $\mod 3$ that the one is superfluous. Let $m=2n+1$. We then $\mod 5$ that $2^{m}+4\equiv 2\cdot 4^{n}-1\equiv 2\cdot (-1)^{n}-1\pmod 5$. This cannot be equal to $0\pmod 5$, which we saw above is inappropriate $2^{m}+4=15b\equiv 0\pmod 5$. $\bullet$ If $c=2$, then $ab=2^k+2$ and $2b-a=1 \ (3)$ and $2a-b=2^m \ (4)$. From the last two, we $b=\frac{2^m+2}{3}$ and $a=\frac{2(2^m+2)}{3}-1$. Substituting in the former and doing deeds, we arrive at $2^{2m+1}+5\cdot 2^m=9\cdot 2^{k}+16$. Because $k\geq m$, we can write $2^{m+1}+5=9\cdot 2^{k-m}+2^{4-m}$. Because $2^{m+1},5,2^{k-m}$ are integers, must also be $2^{4-m}$ an integer. So, $m\leq 4$. By checking the cases (all very simple), we also get the solutions $\boxed{(a,b,c)=(3,2,2),(11,6,2)}$. Therefore, $\boxed{(a,b,c)=(2,2,2),(3,2,2),(11,6,2),(7,5,3)}$, together with all their permutations, they are all solutions to the problem.

Surprisingly the way I reduced the problem into the last two cases makes the argument much harder to see and read compared to not making the $a \to 2a+1$ substitution. Sometimes less methodical is better. Let $ab - c = 2^x$, $bc - a = 2^y$, and $ca - b = 2^z$. First Case: Suppose two of $x, y, z$ are equal, say $x$ and $y$. Then it follows that \[ab-c-bc+a = (a-c)(b-1) = 0.\]If $b = 1$, then $a-c$ and $c-a$ are both powers of $2$, contradiction. Then $a = c$, so $ab-a$ is a power of $2$, meaning that $a$ and $b-1$ are both powers of $2$. When $b = 2$, $a^2$ and $a^2-2$ are powers of $2$ that differ by $2$, hence $a = 2$, yielding the solution $(2, 2, 2)$. When $b \geq 3$, it follows that $a^2 -b = 1$, so $a^2 = b+1$ is a power of two as well as $b-1$, implying $b=3$. This yields $a=2$ and the solution $(2, 3, 2)$ (and cyclic permutations). This exhausts the first case. Second Case: We may assume henceforth that $x, y, z$ are pairwise distinct. Now, suppose that $a, b, c$ are all even, and assume WLOG $1 \leq \nu_2(a) \leq \nu _2(b) \leq \nu_2(c)$. Then, $y = \nu_2(bc-a) = \nu_2(a)$ as \[\nu_2(bc) = \nu_2(b) + \nu_2(c) \geq \nu_2(b) + 1 > \nu_2(a)\]and similarly $z = \nu_2(b)$. Let $a = 2^y k$ and $b = 2^z \ell$ for odd $k, \ell$. By adding the second two equations, we obtain \[(c-1)(a+b) = 2^y + 2^z \iff (c-1)(k+2^{z-y} \ell) = 2^{z-y} + 1.\]Clearly equality can only occur when $k = \ell = 1$ and $c = 2$, so $a = 2^y$ and $b = 2^z$. Hence $2^{y+z} - c = 2^x$, implying $c = 2^{y+z} - 2^x$. On the other hand, the second equation now yields \[2^z(2^{y+z} - 2^x) = 2^{y+1}\]implying $2^{y+z} - 2^x$ is a power of two, thus $y+z = x+1$ and $z+x = y+1$. Thus $z = 1$, and also $y = 1$ by using the third equation, contradicting distinctness. Third Case: If $a, b$ are odd and $c$ is even, then all three cyclic quantities $ab-c$ are odd, so they all equal $1$. Hence $ab+bc+ca = a+b+c+3$ by adding the equations; a quick bound and finite case check yields no solutions. Because this solution is already so long, this part is omitted for length. Fourth Case: If $a, b$ are even and $c$ is odd, then $ab-c=1$. The other two equations read \begin{align*} a^2b-a-b &= 2^y \\ ab^2-a-b &= 2^z. \end{align*}This means that one of the two expressions must divide the other. Say $a^2b -a-b \mid ab^2 -a-b$; then \[a^2b-a-b \mid ab(b-a).\]Let $a = dm$, $b = dn$, where $\gcd(m, n) = 1$, so \[d^2 m^2 n - m-n \mid d^2 mn(n-m).\]As $m$ and $n$ are relatively prime, the LHS is relatively prime to both $m$ and $n$, thus \[d^2 m^2 n \leq d^2(n-m) + (m+n) < 2nd^2.\]So follows $m^2 < 2$, thus $m=1$, and $a$ divides $b$. But then letting $b=da$ yields \[da^3 - a - da = 2^y\]so $a$, $da^2 - d - 1$, $d^2a^2-d-1$ are all powers of $2$. Furthermore, $da^2 - d -1 \mid a^2(d-1)$ by looking at the second and third expressions. If this quotient is at least $2$, then \[a^2 d + a^2 \leq 2d+2 \iff a^2 \leq 2\]contradicting $a$ even. So $a^2 = d+1$ precisely and $d(d+1) - (d+1) = d^2 - 1$ is a power of two. So $d = 3$ is uniquely fixed and $a=2$, yielding the solution $(2, 6, 11)$. Fifth Case: We finish by checking the case where $a, b, c$ are all odd. Here, we substitute $2a+1$ for $a$ and so on, so the equations read \begin{align*} 2ab + a+b-c &= 2^{x-1} \\ 2bc + b+c-a &= 2^{y-1} \\ 2ac + a+c-b &= 2^{z-1}. \end{align*}Again, assume $x<y<z$. Adding the last two equations, we have \[2c(a+b+1) = 2^{y-1} + 2^{z-1}.\]Observe that $2^{y-1} + 2^{z-1}$ is even and greater than $2$, implying that $x, y, z > 1$ by symmetry. Hence, $a, b, c$ are all even, or exactly two of them are odd. If $a, b, c$ are all even, then the previous display implies $\nu_2(c) = y-2$. Similarly, we have $\nu_2(b) = x-2$ and $\nu_2(a) = x-2$. But now, the first two equations imply \[2^{x-1}(2^{x-2} + 2^{y-2} + 1) \leq 2b(a+c+1) = 2^{x-1}+2^{y-1}\]so $2^{x+y-3} \leq 2^{y-1}$ or $x \leq 2$, so $x = 2$. But when $x=2$, the equation reads $2^{y-1} + 4 \leq 2^{y-1} + 2$, contradiction. If $a, b$ are odd and $c$ is even, then we can assume $y<z$ by symmetry, and again $\nu_2(c) = y-2$. But then $\nu_2(2bc) = y-1$, and if $b>1$, this is at least $3\cdot 2^{y-1}$, ergo $a$ is the largest of $a, b, c$. Then we also have $\nu_2(a+1) + \nu_2(c-a) = \operatorname{min}(x-2, z-2)$ by adding first and third equation. As $c-a$ is odd, we have $a+1 > 2^{x-2}$ or $a+1 > 2^{z-2}$. In the first case, $2ab+a+b-c > 2ab+b > 3(2^{x-2} -1) + 3 > 2^{x-1}$, contradiction. Similarly, in the third case, we obtain a contradiction too. Thus $b = 1$, and we need $3a-c + 1$, $3c-a+1$, $2ac+a+c-1$ all powers of $2$. Then $3c-a+1 \mid 4(a-c)$ which has $\nu_2$ precisely two, hence $3c-a+1 \in \{2, 4\}$. Checking the remaining cases yields $(a, b, c) = (2, 1, 3)$ only, which yields the final solution pair $(3, 5, 7)$. In total, the solutions are $(a, b, c) = (2, 2, 2), (2, 3, 2), (2, 6, 11), (3, 5, 7)$ and cyclic permutations.

Bruh the casework The solutions are $\boxed{(2, 2, 2), (2, 2, 3), (3, 5, 7), (2,6,11)},$ and the permutations of these solutions.