If $c=0$ then obviously $a=1$ and $b=0$ Let $c>0$ $mod2$ gives $a>1$ If $p=2$ then $RHS\equiv 1\not\equiv 0\equiv LHS (mod 2)$ contradiction.So $p>=3$ If $b=0$ then we have: If $c=1$ we have the solutionw on the form $(a,b,c,p):(a,0,1,2^a-3)$ with $p=prime$ If $c=odd>1$ then $2^a=(p+3)[(p+2)^{c-1}-...+1]$ wich gives contradiction as the second factor is odd. If $c=even$ then $mod4$ gives $a=1$ contradiction. If $p=3$ we have $2^a3^b=5^c+1$ with $a,b,c>0$ Taking $mod4,9,7$ we have $a=b=c=1$ Now we have to solve the folowing: Find all positive integers $a,b,c$ and prime $p\geqslant 5$ such that $2^a p^b=(p+2)^c+1$ If $c=odd$ then: $2^ap^b=(p+3)[(p+2)^{c-1}-...+1]$ From here we can easily see that $2^a=p+3$ If $k|c$ and $1\neq k\neq c$ then $p|(p+2)^k+1|LHS$ but then by ZSIGMONDY we have contradiction.So $c=1,prime$ If $c=1$ then $p=3$ contradictin as $p>3$ So $c=prime=q$ $(p+2)^q+1\equiv 0(modp)\Rightarrow 2^q+\equiv -1(modp)\Rightarrow 2^{2q}\equiv 1(modp)$ As $p>=5$ we can not have $2^{2or1}\equiv 1(modp)$.So $2q$ is the order of $2(modp)$. This means that $2q|p-1$ but as $U_2(2q)=1$ and $U_2(p-1)=2$ we have that $2q|\frac{p-1}{2}$ $(\frac{2}{p})\equiv 2^{\frac{p-1}{2}}\equiv 1(modp)\Rightarrow (\frac{2}{p})=1\Rightarrow p\equiv 1,7(mod8)$ But $p=2^a-3=5(mod8)$ So no solution if $c=odd$ If $c=even$ Then $mod4$ gives $a=1$ and so we have $2 p^b=(p+2)^{2c_1}+1$ $p|(p+2)^{2c_1}+1=x^2+1$ so $p\equiv 1(mod4)$ Let $q=prime$ such that$q|p+2$ and $q\equiv 3(mod4)$ then: If $b=odd$ we will have: $2(-2)^b\equiv 1(modq)\Rightarrow q|2^{b+1}+1=x^2+1$ contradiction.So $b=even$ For $b=even$ the last one became $2 p^{2b_1}=(p+2)^{2c_1}+1$ Taking $mod3$ we have $p=2(mod3)$ $p|LHS=x^2+1$ so $p=1(mod4)$ $[(p+2)^{c_1},p^{b_1}]=(x_{n'},y_{n'})$ is a solution of the negative pell eqution $x^2-2y^2=-1$ We can find that: $x_{n+2}=6x_{n+1}-x_n$ with $x_1=7$,$x_2=41$ $y_{n+2}=6y_{n+1}-y_n$ with $y_1=5$,$y_2=29$ If $b_1=odd$ we will prove that $p=5$ Taking the first sequens $mod3$ gives $1,2,2,1$ and as $(p+2)^{c_1}\equiv 1(mod3)$ we have $n'\equiv 1,0(mod4)$ Taking the second sequens $mod3$ gives $2,2,1,1$ and as $p^{b_1}\equiv 2(mod3)$ we have $n'\equiv 1,2(mod4)$ So $n'\equiv 1(mod4)$ Taking the first sequens $mod4$ gives $3,1$ and as $n'=odd$ we must have: $3\equiv (p+2)^{c_1}(mod4)\Rightarrow 3\equiv 3^{c_1}(mod4)\Rightarrow c_1=odd$ If $p\equiv 1(mod5)\Rightarrow LHS\equiv 2\not\equiv 0\equiv RHS(mod5)$ contradiction If $p\equiv 2(mod5)\Rightarrow LHS\equiv 3\not\equiv 2\equiv RHS(mod5)$ contradiction If $p\equiv 3(mod5)\Rightarrow LHS\equiv 2\not\equiv 1\equiv RHS(mod5)$ contradiction If $p\equiv 4(mod5)$ then: Taking the first sequens $mod5$ gives $2,1,4,3,4,1$ and as $(p+2)^{c_1}\equiv 1(mod5)$ we get $n'\equiv 0,2(mod6)$ but $n'=odd$ contradiction. So $p=5$ and we have $2*5^{2b_1}=7^{2c_1}+1$ If $b_1=1$ then $c_1=1$ If $b_1>1$ then: $mod 125$ gives $5|c_1$ $mod11$ gives $5|b_1$ $mod7$ gives $b_1=1(mod3)$ And finaly $mod31$ gives $RHS\equiv 25,5,2\not\equiv 10\equiv LHS(mod31)$ contradiction. If $b_1=even$ then we have the follwing:$2 p^{4b_2}=(p+2)^{2c_1}+1$ Taking the first sequens $mod3$ gives $1,2,2,1$ and as $(p+2)^{c_1}\equiv 1(mod3)$ we have $n'\equiv 1,0(mod4)$ Taking the second sequens $mod3$ gives $2,2,1,1$ and as $p^{b_1}\equiv 1(mod3)$ we have $n'\equiv 3,0(mod4)$ So $n'\equiv 0(mod4)$ Taking the first sequens $mod4$ gives $3,1$ and as $n'=even$ we must have: $1\equiv (p+2)^{c_1}(mod4)\Rightarrow 1\equiv 3^{c_1}(mod4)\Rightarrow c_1=even$ And now from Mathematical Danube Competition 2017, Juniors P4 Determine all triples of positive integers $(x,y,z)$ such that $x^4+y^4 =2z^2$ and $x$ and $y$ are relatively prime

I hate this... (yeah thats like the only comment i have about this, i'm not gonna share a motivational history or something like that, duh!. I took like 2.5 hrs on this thing so like i wanna die aaaaaa) Case 1: $c<0$ Then let $c=-z$ and re-write the diophantine as $(p+2)^z2^ap^b=(p+2)^z+1$ and clearly $a,b$ cant both be non-negative so lets begin with the casework. Case 1.1: $a,b \le 0$ Then let $a=-x$ and $b=-y$ and re-write as $(p+2)^z=((p+2)^z+1)2^xp^y \ge (p+2)^z+1$ which is imposible by size reazons. Case 1.2: $a \le 0$ and $b \ge 0$ Then re-write as $(p+2)^zp^b=((p+2)^z+1)2^x$ and if $x \ne 0$ then using mod $2$ we get $p=2$ and then u get $2^{2z+b-x}-2^{2z}=1$ which is not possible so $x=0$ but if $x=0$ then by mod $p+2$ u get $p+2 \mid 1$ which is not possible. Case 1.3: $a \ge 0$ and $b \le 0$ Then re-write as $(p+2)^z2^a=((p+2)^z+1)p^y$ and now if $y \ne 0$ then by mod $p$ u get $p=2$ but then u get a subcase from the previous case and we've seen it has no sols, so this one also doesnt, now if $y=0$ then use mod $p+2$ to see there are no solutions. Case 2: $c=0$ Here u dont really need such casework, small work gives that the set of solutions here is $(a,b,c,p)=(a,1-a,0,2), (1,0,0,p)$ Case 3: $c>0$ First this means that $a,b \ge 0$ and now even more cases aaaaaa Case 3.1: $c$ is odd. Then we divide into 2 subcases that seem to make no sense but they do have sense actualy. Case 3.1.1: $p$ is a prime of the form $2^n-3$ Then our diophantine is $2^a(2^n-3)^b=(2^n-1)^c+1$ and from here u can get $a=n$ which means that $2^a(2^a-3)^b=(2^a-1)^c+1$ and now by mod $2^a-2$ u get $b$ even and by mod $2^a-1$ u get $2^a-1 \mid 2^b-1$ which means $a \mid b$ and this means that u have 2 perfect powers consecutive, and by catalan's conjeture u get a contradiction if $c \ne 1$ so now if $c=1$ u get $(2^a-3)^b=1$ which means $b=0$ so the only sol here is $(a,b,c,p)=(a,0,1,p')$ where $p'$ denotes a prime of the form $2^a-3$. Case 3.1.2: $p+3$ is not a power of $2$ Then let $q$ an odd prime such that $q \mid p+3$ and now by mod $q$ u get $q=p$ which means $p \mid p+3$ which means $p=3$ so our equation is $2^a3^b=5^c+1$ and now if $c>1$ then by zsigmondy there exists a prime factor $q$ of $5^c+1$ that its not $2$ or $3$ but u get a contradiction by mod $q$ hence $c=1$ which means $2^a3^b=6$ and from here u get that the only sol on this case is $(a,b,c,p)=(1,1,1,3)$ Case 3.2: $c$ is even. Let $c=2^n \cdot m$ where $m$ is odd, then using orders u get $2^{n+1} \mid p-1$, now by mod $4$ u get that $a=1$ now our diophantine is $2p^b=(p+2)^c+1$ or $2(p^b-1)=(p+2)^c-1$ and now take $v_2$ on both sides using LTE to get $v_2(p+1)+v_2(p-1)+v_2(b)=v_2(p+1)+v_2(p+3)+n-1$ and now note that $v_2(p+3)$ is $2$ and $v_2(p-1) \ge n+1$ and with this u have $n+1 \ge v_2(b)+n+1 \ge n+2$ but there is no way this happens, so u get a contradiction, hence no sol on this case. Now that ALL the cases are killed we get the following set of solutions (casework got me tired aaaa) $$\boxed{(a,b,c,p)=(a,1-a,0,2), (1,0,0,p), (a,0,1,p'=2^a-3), (1,1,1,3)}$$And with this, we are done (aaaaaa for the sake of my sanity i need a rest)

@above, a=1,b=2,c=2,p=5 works. BTW, the problem asks us to find positive integers, therefore you don't need to consider a,b,c<0.

If c is even, then a=1. By Zsigmonndy's theorem, there is a prime that divides (p+2)^c+1 and does not divide (p+2)^i+1 for all i<c. If c is not a power of 2, then it is easty to reach a contradiction. If c is a power of 2, then p=1(mod 2c) and v2(p-1)=v2(c)+1. Now if b>=c+1 we will reach a contradiction since p-1>=6c. Therefore p=2c+1 and b=c, and the rest is easy. If c is odd, WLOG p+3 is a power of 2, then 2^a=p+3, thus (p+3)p^b=(p+2)^c+1. Similarly, using Zsigmondy, we must have p=1(mod 2c), and a simple bound gives us b=c-1 and p=2c+1. The rest is also easy.

Here is a very elementary approach, and probably the intended one because China loves problems that test students' number sense. Nice bounding in two directions problem. The answer is $(a,b,c,p)=(1,1,1,3), (1,1,2,5)$ only. Suppose $\nu_2(c)=k$, then $(p+2)^{2^k}+1\mid (p+2)^c+1$ Claim: If $k\ge 2$, there exists no solution to $(p+2)^{2^k}+1=2^ap^b$ Proof: By mod 4, $a=1$ Since $p\mid 2^{2^k}+1, ord_p(2)=2^{k+1}$ so $p\equiv 1(\bmod\; 2^{k+1}) $, so $p\ge 17$ Now, $1<\frac{(p+2)^{2^k}}{p^b}<2$. Notice $1\le \frac{(p+2)^{2^k}}{p^{2^k}}=(1+\frac 2p)^{2^k} \le (1+\frac 1p)^{2^{k+1}} \le (1+\frac 1p)^p \le e$ (this is the key step) Therefore, if $b\ne 2^k$ getting a size contradiction is easy. Now, $b=2^k$. We have $2(p^{2^k}-1)=(p+2)^{2^k}-1$, and LTE works. Back to the original problem. We've already ruled out the case where $4\mid c$. You find $\nu_2(c)=0 \rightarrow p=3$ by settling the case $c=1$ and very basic bounding, $\nu_2(c)=1 \rightarrow p=5$ by settling the case $c=2$ and very basic bounding and it's easy to finish with LTE to see $(a,b,c,p)$ can only be $(1,1,1,3)$ and $(1,1,2,5)$ I wonder what bora_olmez's solution that uses $\mathbb{Z}[i]$ and $\mathbb{Z}[\sqrt{2i}]$. It should be posted in all the solution's honor.

@above yes, the (1+1/p)^p<e is exactly the key step for this problem, that's what I did in order to bound b and p, they should not be too far from c and we now have a single variable equation

My 300th post!!! Solution Notice that $2\mid 2^ap^b,$ we have $p\equiv 1\pmod 2.$ Let $c=2^{\theta}r$ where $2\nmid r,$ then $(p+2)^{2^{\theta}}+1\mid 2^ap^b.$ Lemma $:$ $(p+2)^{2^{\theta}}+1$ is divisible by $p.$ If it is not true$,$ $(p+2)^{2^{\theta}}+1=2^{\beta},$ then $\beta >2.$ Therefore $(p+2)^{2^{\theta}}+1\equiv 0\pmod 4,\theta =0,c=r,$ then $p+3=2^{\beta},$ therefore $p\equiv 5\pmod 8,$ so that $\left(\frac{-2}p\right)=-1.$ However $2^{r+1}\equiv -2\pmod p,$ contradiction! Therefore $(p+2)^{2^{\theta}}+1$ is divisible by $p.$ If $r\geq 3,$ let $N=\frac{(p+2)^{2^{\theta}r}+1}{(p+2)^{2^{\theta}}+1},$ then $N=\sum\limits_{j=0}^{r-1}(p+2)^{j\cdot 2^{\theta}}(-1)^j\equiv r\equiv 1\pmod 2,$ therefore $2\nmid N.$ From $N\mid 2^ap^b,$ $N=p^k,$ where $k\in\mathbb N.$ Using LTE Lemma$,$ $v_p(N)=v_p(r),$ therefore $N\leq r.$ Let $x=(p+2)^{2^{\theta}},$ then $x\geq 5,$ therefore $N=\frac{x^r+1}{x+1}\geqslant x^{r-1}-x^{r-2}\geqslant\frac 12x^{r-1}>r,$ contradiction! Now $r=1.$ If $\theta =0,c=1,$ then $p+3=2^ap^b\equiv 0\pmod p,$ so $p=3,a=b=1,(a,b,c,p)=(1,1,1,3).$ Consider $\theta\geq 1,$ then $2^ap^b=(p+2)^{2^{\theta}}+1\equiv 2\pmod 4,$ therefore $a=1,2p^b=(p+2)^{2^{\theta}}+1\cdots (*).$ If $\theta =1,$ $2^ap^b=p^2+4p+5,$ therefore $p=5,$ we can get $b=2,(a,b,c,p)=(1,2,2,5).$ Now let $\theta\geq 2,$ then $2^{2^{\theta}}\equiv -1\pmod p,2^{2^{\theta +1}}\equiv 1\pmod p,$ so $\delta _p(2)=2^{\theta +1},2^{\theta+1}\mid (p-1).$ Therefore $p\equiv 1\pmod {16}.$ From $(*)$ we have $2(-1)^b\equiv 2\pmod {p+1},$ so $2\mid b.$ We can write $(*)$ as $2(p^b-1)=(p+2)^{2^{\theta}}-1.$ Using LTE Lemma$,$ $v_2\left((p+2)^{2^{\theta}}-1\right)=v_2\left((p+2)^2-1\right) +\theta -1=\theta +2.$ However$,$ $\theta +1=v_2(p^b+1)\geqslant v_2\left(p^2-1\right)\geqslant v_2(p-1)+v_2(p+1)\geqslant\theta +2,$ contradiction! Therefore $\boxed{(a,b,c,p)=(1,1,1,3),(1,2,2,5)}.\blacksquare$