$(x,y,z,w) = (1,1,1,0)$, $(2,2,1,1,)$, $(1,0,0,0)$ or $(3,0,0,1)$ If $x = 0$, then we get odd-odd=odd, contradiction If $x\geq 3$, $y \geq 1$, then mod 3, we get $z$ is odd, mod 8, we get $z$ is even, contradiction. if $y \geq 1$, $2\geq x \geq 1$ then [ if $x=1$, then if $z\geq 1$, $w \geq 1$[ mod 7, we find that $y=4$ (mod 6), which means that $y$ is even. mod 5, we find that $y=1$ mod 4, which means that $y$ is odd, contradiction. ]else if $z = 0$, $w \geq 0$[ If $y=0$, then we get the solution $(1,0,0,0)$. we check that $y=1$ dosen't work, so mod 9, we get a contradiction. ]else if $z > 0$, $w = 0$[ If $y \leq 1$, we check to find the solution $(1,1,1,0)$. Otherwise, mod 9, and we find that $z$ is divisible by 3. However since $5^{3k} +1$ is divisible by 7, we get a contradiction. ] if $x=2$, then [ we find the solution $(2,2,1,1)$. Mod 7 or mod 5, we find that y is even. Let $y= 2g$. If y is larger than 4 then $5^z7^w = 4*3^y -1 = (2*3^g-1)(2*3^g+1)$. We find that $(2*3^g-1)$ must equal to $5^z$ or $7^w$, where $g>1$. We have already proven that this is impossible. (this is equivalent to the previous two cases.) ] ] Now, consider the case when $y=0$. Mod 7, we find that $x$ is divisible by 3. If $z \geq 1$, then mod 5, we find that $x$ is divisible by 4. However, since $2^{2k}-1$ is divisible by 3, thus, contradiction. Finally, we are left with the case that $y=0$, $z=0$. If $x\leq3$, then we check and find the solution $(3,0,0,1)$, and if $x>3$, mod 16 would yield a contradiction. Q.E.D. P.S. There got to be a better solution!

i have the other solution first we easy to prove that x <= 2 then we have two case: if x=1 then easu to prove that the equation hasn't root (use modular) if x=2 then we mast use Pell's equation and we have the result (for the case two it's two long and i am very lazy to write it)

maybe I'm blind, but how do you particularly reach pell's equation in here. And even if u do I don't think it gets easier than the ordinary solution.

orl wrote: Find all nonnegative integer solutions $ (x,y,z,w)$ of the equation \[ 2^x\cdot3^y - 5^z\cdot7^w = 1. \] It is clear that $ x\geq 1$,because otherwise left side of equation is even. Suppose $ y = 0$: Then if $ z\geq 1$,we conclude that $ 2^{x}\equiv 1$ mod $ 5$,but it means that $ 4|x$,hence $ 3|2^{x} - 1$,so $ 3|5^{z}\cdot 7^{w}$,contradiction. Therefore,$ z = 0$ and $ 2^{x} = 1 + 7^{w}$. Now if $ x\geq 4$,it follows that $ 7^{w}\equiv 1$ mod $ 16$,but as one can check,it is impossible. Therefore,$ x\leq 3$.Checking cases $ x = 1,3$,it gives that $ \boxed{(1,0,0,0)}$ and $ \boxed{(3,0,0,1)}$ are solutions. Now we can assume $ y > 0$.Consider two cases: 1-If $ x = 1$: Then considering our equation modulo $ 3$,we conclude that $ z$ is odd,thus $ z\geq 1$,and after that modulo $ 5$,which gives $ y\equiv 1$ mod $ 4$. Suppose $ w > 0$,it follows that $ 2\cdot 3^{y}\equiv 1$ mod $ 7$,but last congruence is impossible,as $ y\equiv 1$ mod $ 4$. Therefore,$ w = 0$,and our equation can be rewritten,as follows: $ 2\cdot 3^{y} - 5^{z} = 1$. As we've observed $ y\equiv 1$ mod $ 4$,hence if $ y = 1$: Then $ z = 1$,a new solution is $ \boxed{(1,1,1,0)}$. If $ y\geq 5$,then $ 5^{z}\equiv - 1$ mod $ 9$,hence $ z = 6k + 3$,or $ 5^3 + 1|5^{z} + 1|2\cdot 3^{y}$,contradiction. 2-If $ x\geq 2$: Consider our equation modulo $ 4$ and modulo $ 3$: $ 7^{w}\equiv - 1$ mod $ 4$,hence $ w$ is odd. $ 5^{z}\equiv - 1$ mod $ 3$,hence $ z$ is odd. So $ w$ and $ z$ are odd,hence $ 2^{x}\cdot 3^{y}\equiv 5^{z}\cdot 7^{w} + 1\equiv 5\cdot 7 + 1\equiv 4$ mod $ 8$. Therefore $ x = 2$ and $ y$ is even. Consider our equation modulo $ 5$ and $ 7$: $ 4\cdot 3^{y}\equiv 1$ mod $ 5$ and $ 7$,hence $ \boxed{y = 12k + 2}$. We can rewrite our equation as: $ (2\cdot 3^{6k + 1} - 1)(2\cdot 3^{6k + 1} + 1) = 5^{z}\cdot 7^{w}$. As $ \gcd(2\cdot 3^{6k + 1} - 1,2\cdot 3^{6k + 1} + 1) = 2$ and $ 7\not{|}2\cdot 3^{6k + 1} - 1$,it follows that $ 2\cdot 3^{6k + 1} + 1 = 7^{w}$ and $ 2\cdot 3^{6k + 1} - 1 = 5^{z}$ But the second equation was considered in first case,hence $ k = 0$ and $ y = 2$. So the last solution is $ \boxed{(2,2,1,1)}$.(all solutions were written in form $ (x,y,z,w)$.

The answer is $(1,0,0,0)$, $(3,0,0,1)$, $(1,1,1,0)$ and $(2,2,1,1)$. Zsigmondy approach First, repeated applications of Zsigmondy's theorem solve the problem immediately in the situation where $abcd = 0$, so from now on we assume $a,b,c,d > 0$. (This is not too difficult even without Zsigmondy, but involves several cases.) Taking modulo $3$ gives $1 + 2^c \equiv 0 \pmod 3$ so it follows $c$ is odd. Taking modulo $8$ gives $2^a \cdot 3^b \equiv 1 + 5 \cdot 7^d \pmod 8$ from which it follows that $a < 3$. So either $a = 1$ or $a = 2$. Taking modulo $5$ gives $1 \equiv 2^a 3^b \equiv 2^{a-b} \pmod 5$, so it follows that $a \equiv b \pmod 4$. Taking modulo $7$ gives $2^a 3^b \equiv 1 \pmod 7$. Since $a$ and $b$ have the same parity, this proves $a \neq 1$ (since otherwise $6M^2 \equiv -1 \pmod 7$). So we are reduced to the case where $a = 2$ and $b \equiv 2 \pmod 4$. Let $b = 2e$. Then a difference of squares gives \[ \left( 2 \cdot 3^e - 1 \right)\left( 2 \cdot 3^e + 1 \right) = 5^c \cdot 7^d. \]Since the two factors are relatively prime and the first is $2 \pmod 3$, it follows $2 \cdot 3^e = 5^c + 1$ and $2 \cdot 3^e = 7^d - 1$. Another application of Zsigmondy gives $c=d=e=1$, so $(a,b,c,d) = (2,2,1,1)$. Longer approach without Zsigmondy Here is a solution without Zsigmondy. It's clear that $a \ge 1$, otherwise the left-hand side is even. The remainder of the solution involves several cases. First, suppose $b=0$. If $c \ge 1$, then modulo $5$ we discover $2^a \equiv 1 \pmod 5$ and hence $4 \mid a$. But then modulo $3$ this gives $-5^c 7^d \equiv 0$, which is a contradiction. Hence assume $c=0$. Then this becomes $2^a-7^d=1$. This implies $1+7^d \equiv 2^a \pmod{16}$, and hence $a \le 3$. Exhausting the possible values of $a=0,1,2,3$ we discover that $(3,0,0,1)$ and $(1,0,0,0)$ are solutions. Henceforth suppose $b > 0$. Taking modulo $3$, we discover that $5^c \equiv -1 \pmod 3$, so $c$ must be odd and in particular not equal to zero. Then, taking modulo $5$ we find that \[ 1 \equiv 2^a3^b \equiv 2^{a-b} \pmod 5. \]Thus, $a \equiv b \pmod 4$. Now we again have several cases. First, suppose $d=0$. Then $2^a3^b = 5^c+1$. Taking modulo $4$, we see that $a=1$ is necessary, so $b \equiv 1 \pmod 4$. Clearly we have a solution $(1,1,1,0)$ here. If $b \ge 2$, however, then taking modulo $9$ gives $5^c \equiv -1 \pmod 9$, which occurs only if $c \equiv 0 \pmod 3$. But then $5^3+1 = 126$ divides $5^c+1 = 2^a3^b$, which is impossible. Now suppose $d \neq 0$ and $a$, $b$ are odd. Then $6M^2 \equiv 1 \pmod 7$, where $M = 2^{\frac{a-1}{2}}3^{\frac{b-1}{2}}$ is an integer. Hence $M^2 \equiv -1 \pmod 7$, but this is not true for any integer $M$. Finally, suppose $b, c, d \neq 0$, and $a=2x$, $b=2y$ are even integers with $x \equiv y \pmod 2$, and that $c$ is odd. Let $M = 2^x 3^y$. We obtain $(M-1)(M+1) = 5^c 7^d$. As $\gcd(M-1, M+1) \le 2$, this can only occur in two situations. In one case, $M-1=5^c$ and $M+1=7^d$. Then $5^c+1 \equiv 2^x 3^y$. We have already discussed this equation; it is valid only when $x=y=1$ and $c=1$, so $(a,b,c,d) = (2,2,1,1)$. In the other case, $M+1=5^c$ and $M-1=7^d$. Taking the first relation modulo $3$, we obtain that $M \equiv 2^c-1 \equiv 1 \pmod 3$. Hence $y=0$, and $x$ is even. Now $2^x+1 = 5^c$ and $2^x-1=7^d$. But if $x$ is even then $3 = 2^2-1 \mid 2^x-1 \mid 7^d$, which is impossible. Hence there are no solutions here. In summary, the only solutions are $(1,0,0,0)$, $(3,0,0,1)$, $(1,1,1,0)$ and $(2,2,1,1)$.

can anyone explain to me how does Zsigmondy work here? doesn't it state,that $ a^n - b^n $ has at least one prime factor that does not divide $ a^k - b^k $ (yes it has some exceptions)? or does this theorem has some corollaries?

First let's make sure we agree on the statement of Zsigmondy, since there are a few variants. Zsig: Let $a < b$ be relatively prime positive integers. Then $a^n-b^n$ has a primitive prime divisor for all $n$, except when $n=2$ and $a+b=2^k$, or $(a,b,n)=(1,2,6)$. $a^n+b^n$ has a primitive prime divisor for all $n$, except when $(a,b,n)=(1,2,3)$. Basically what this means is that once we pin down a Diophantine equation to something like $c^n \pm 1 = p_1^{e_1} p_2^{e_2} \dots p_k^{e_k}$ for some fixed primes, we are immediately done modulo small cases. For a concrete example, take the Diophantine equation \[ 19^n - 1 = 2^a3^b5^c7^d11^e13^f17^g. \] Let $n$ run from $1$ to $\infty$. You can check by hand that all the primes appear by the time you reach $n=16$. So for $n \ge 17$ there are no solutions by Zsig, since there must be a prime $p$ dividing $19^n-1$ that we haven't seen before.

v_Enhance wrote: First let's make sure we agree on the statement of Zsigmondy, since there are a few variants. Zsig: Let $a < b$ be relatively prime positive integers. Then $a^n-b^n$ has a primitive prime divisor for all $n$, except when $n=2$ and $a+b=2^k$, or $(a,b,n)=(1,2,6)$. $a^n+b^n$ has a primitive prime divisor for all $n$, except when $(a,b,n)=(1,2,3)$. Basically what this means is that once we pin down a Diophantine equation to something like $c^n \pm 1 = p_1^{e_1} p_2^{e_2} \dots p_k^{e_k}$ for some fixed primes, we are immediately done modulo small cases. For a concrete example, take the Diophantine equation \[ 19^n - 1 = 2^a3^b5^c7^d11^e13^f17^g. \] Let $n$ run from $1$ to $\infty$. You can check by hand that all the primes appear by the time you reach $n=16$. So for $n \ge 17$ there are no solutions by Zsig, since there must be a prime $p$ dividing $19^n-1$ that we haven't seen before. thank you so much,that's clear after all (:

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=56&t=587398&p=3481126#p3481126 My solution is very straightforward and it doesn't take many lines to write it. However, I want to point out that there is also a solution using Stormer's Theorem. It will take me a great amount of time to write it, but it's completely mechanic.

The answers are $(1,0,0,0)$, $(1,1,1,0)$, $(2,2,1,1)$, $(3,0,0,1)$. Claim. If $x,y,z,w>0$, then $(x,y,z,w)=(2,2,1,1)$. Proof. First assume $x\ge3$. Taking modulo $8$, we have $5^z7^w\equiv-1\pmod8$, however $5^z\in\{1,5\}$ and $7^w\in\{1,7\}$, so we must have $(5^z,7^w)=(1,7)$. It follows that $z$ is even and $w$ is odd. However taking modulo $6$, we have $-(-1)^z\equiv1\pmod6$, so $z$ is odd, contradiction. Hence we may take the following two cases: Assume $x=1$. Then $5^z7^w=2\cdot3^b-1$, so $2\cdot3^y\equiv1\pmod{35}$. This modulo $5$ gives $y\equiv1\pmod4$, and this modulo $7$ gives $y\equiv4\pmod6$, absurd. Assume $x=2$. Then $5^z7^w=4\cdot3^y-1$, so $4\cdot3^y\equiv1\pmod{35}$. Modulo $5$ gives $y$ is even, so \[5^z7^w=(2\cdot3^{y/2}-1)(2\cdot3^{y+2}+1).\]The two terms are relatively prime, so they must be $5^z$ and $7^w$ in some order. That is, either $2\cdot3^{y/2}=5^z+1$ or $5^z-1$. By Zsigmondy theorem, $z\le2$, and it is easy to check that only $(2,2,1,1)$ works. This exhausts all cases, thus proving the claim. $\blacksquare$ To finish, note the following: If $x=0$, then the left-hand expression is even, so no solutions. If $y=0$, then $5^z7^w=2^x-1$, so by Zsigmondy theorem we need $x\le4$ or $x=6$, for which only $(1,0,0,0)$ and $(3,0,0,1)$ work. If $z=0$, then $2^x3^y=7^w+1$, but the right-hand expression is always $2\pmod3$, so $y=0$, and by Mih\u ailescu theorem only $(1,0,0,0)$ and $(3,0,0,1)$ work. If $w=0$, then $2^x3^y=5^z+1$, so by Zsigmondy theorem, $z\le2$, thus only $(1,0,0,0)$ and $(1,1,1,0)$ work. This completes the proof.

Eyed has a solution; I think it's really well-written: Eyed wrote: The only solution is \[(3,0,0,1), (2,2,1,1), (1,1,1,0), (1,0,0,0)\] Case 1: $a\geq 3, b\geq 1$ If $a\geq3$ and $b\geq 1$, then taking $\mod 24$, we get $5^{c}7^{d}\equiv -1\mod 24$. Taking $\mod 8$ we get $5^{c}7^{d}\equiv -1\mod 8$, which implies $c$ is even and $d$ is odd. Then, if $c$ is even, $5^{c}\equiv 1\mod 24$, and if $d$ is odd, $7^{d}\equiv 7\mod 24$, so $5^{c}7^{d}\equiv 7\mod 24\not\equiv-1\mod 24$, a contradiction. There are no solutions for this case. Case 2: $a\geq 3$, $b = 0$ We get $2^{a} - 5^{c}7^{d} =1$. If $c\geq 1$, then taking $\mod 5$ results in $a$ even, taking $\mod 8$ results in $c$ even, and taking $\mod 3$ gives us $1-1\equiv 1\mod3$, a contradiction. Otherwise, assume $c = 1$. Then, we get $2^{a} - 7^{d} = 1$. If $d\geq 2$, then by Catalan's conjecture there are no solutions. Thus, we only need to check $d = 0,1$, in which $(a,d) = (3,1)$ works, which is the only solution for this case. Case 3: $b \geq1$ and $a = 2$. We have \[4\cdot 3^{b} -1 = 5^{c}7^{d}\]Taking $\mod 3$ results in $c$ is odd, so $c\geq 1$. Taking $\mod 5$ results in $b\equiv2\mod4$. Taking $\mod 4$ also reveals that $d$ is odd, so $d\geq 1$. Taking $\mod7$ results in $b\equiv2\mod 6$, so $b\equiv2\mod12$. Then, we let $b = 2b_{1}$. Factoring, we get \[(2\cdot3^{b_{1}} - 1)(2\cdot3^{b_{1}} + 1) = 5^{c}7^{d}\]Either $b_{1}\equiv 1\mod 12$ or $b_{1}\equiv 7\mod 12$. In both cases, we can't have one factor be divisible by both $7$ and $5$, so one factor is equal to $5^{c}$ and the other $7^{d}$. If $2\cdot 3^{b_{1}} + 1 = 5^{c}$, taking $\mod 4$ results in no solutions. If $2\cdot 3^{b_{1}} + 1= 7^{d}$, then \[b_{1} = v_{3}(2\cdot 3^{b_{1}}) = v_{3}(7^{d} - 1) = v_{3}(6) + v_{3}(d) = 1 + v_{3}(d)\]so $v_{3}(d) = b_{1} - 1$, and $d\geq 3^{b_{1} - 1}$. It is easy to see that, for any $b_{1}>1$, the RHS is much larger than the LHS. The only solution is therefore $b_{1} = 1, d = 1$. For this case, the only solution is $b = 2, c = 1, d = 1$. Case 4: $b \geq 1, a= 1$ Our equation becomes \[2\cdot 3^{b}- 5^{c}7^{d} = 1\]Taking $\mod 3$, we get that $c$ is odd, so $c\geq 1$. Then, taking $\mod 5$, we get that $b$ is odd. If $d\geq 1$, then taking $\mod 7$ we have $c\equiv 4\mod 6$, a contradiction, so $d = 0$. Our equation now becomes $2\cdot3^{b} = 5^{c} + 1$. Then, we have $b = v_{3}(5^{c} + 1) = v_{3}(6) + v_{3}(c)$, so $v_{3}(c) = b-1 \Rightarrow c\geq 3^{b-1}$. Then, for all $b\geq 1$, the RHS would be much greater than the LHS. Thus, the only solution for this case is $b = 1, c = 1, d = 0$. Case 5: $b\geq 1, a = 0$ Our equation becomes \[3^{b} - 5^{c}7^{d} = 1\] Taking $\mod 3$ we get $c$ is odd, so $c\geq 1$. Then, taking $\mod 5$, we have $b\equiv 0\mod 4$. Let $b = 2b_{1}$. Then we get \[(3^{b_{1}} - 1)(3^{b_{1}} + 1) = 5^{c}7^{d}\]If one of the factors is divisible by both $5$ and $7$, then the other factor is divisible by neither, so the other factor must be $1$, which is impossible. Thus, assume each factor is a power of $5$ or $7$. If $3^{b_{1}} - 1 = 5^{c}$, for any $c\geq2$ this is impossible by Catalan's conjecture, and for $c = 0,1$ we can check it also doesn't work. If $3^{b_{1}}-1 = 7^{d}$, again if $d\geq 2$ this is impossible by Catalan's, and if $d = 0,1$ we can again check it does not work. There are no solutions for this case. Case 6: $b<1, a<3$ We can hand check all the values of $a$ and $b$ then, which are $(a,b) = (2,0), (1,0), (0,0)$. The only values that work are $(a,b,c,d) = (1,0,0,0)$. Now, since we've covered all the cases, we can assure that the only solutions are $(a,b,c,d) = $ \[(3,0,0,1), (2,2,1,1), (1,1,1,0), (1,0,0,0)\]

First we deal with the edge cases where one of the variables is 0. Suppose $a=0$, then we have $3^b-1=5^c7^d$. We can check $b\in [0, 6]$ don't yield any solutions, and for $b\ge 7$, we can apply Zsigmondy to get a contradiction. Next, suppose $b=0$ and $a>0$, so $2^a-1=5^c7^d$. We see $a=1, c=d=0$ and $a=3, c=0, d=1$ work, and $a=2, 4, 6$ don't work, and for all other $a$ we can apply Zsigmondy to get a contradiction. Next, suppose $c=0$ and $a, b>0$, so $2^a3^b=7^d+1$. Taking mod 3 immediately yields a contradiction. Finally, suppose $d=0$ and $a, b, c>0$, so $2^a3^b=5^c+1$. We see $c=1, a=b=1$ works, and for $c\ge 2$ we can apply Zsigmondy to get a contradiction. Putting this all together, we find that the solutions we get when one of the variables is 0 are $(1, 0, 0, 0), (3, 0, 0, 1), (1, 1, 1, 0)$. Now, suppose $a, b, c, d\ge 1$. Taking mod 3 we get $c$ is odd. We proceed by splitting into cases based on the value of $a$. If $a\ge 3$, then taking mod 4 we get $d$ is odd, but then taking mod 8 we get $5^c\equiv 1\pmod{8}$, which is a contradiction since $c$ is odd. If $a=1$, then taking mod 5 we get $b\equiv 1\pmod{4}$, but then taking mod 7 we get $3^b\equiv 4\pmod{7}$, which is a contradiction since $b$ is odd. Finally, we deal with $a=2$; first note $(2, 2, 1, 1)$ is a solution. Now, taking mod 5 we get $b$ even, so set $b=2b_1$, and we have the equation $(2\cdot 3^{b_1}+1)(2\cdot 3^{b_1}-1)=5^c7^d$. The two factors on the LHS are relatively prime, so we either have $7^d+1=2\cdot 3^{b_1}$, or $7^d-1=2\cdot 3^{b_1}$. The former case dies mod 3. For the latter, note $d=1$ gives us the solution we already have, and $d\ge 2$ gets killed by Zsigmondy. Hence, putting everything together, our solutions for $(a, b, c, d)$ are $(1, 0, 0, 0), (3, 0, 0, 1), (1, 1, 1, 0), (2, 2, 1, 1)$.

Mod bash because I don't like Zsigmondy lol (jk I'm just dumb and forgot): We just consider many cases which I won't go into too much detail. If $b = d = 0$, notice that $a = 1$ a solution when $c = 0$ so $(1, 0, 0, 0)$ is a solution, but when $a \ge 2$ mod $4$ gives no solutions. If $b = c = 0$, notice by mod $64$ that there does not exist an integer $d$ where $7^d \equiv -1 \pmod {64}$, and also notice by mod $7$ that $a \equiv 0 \pmod 3$. When $a = 0$ we get the solution we already had. When $a = 3$ we get $d = 1$ so $(3, 0, 0, 1)$ is a solution, but when $a \ge 6$ we get mod $64$ contradiction. If $a = 0$ then mod $2$ gives no solutions. If $b = 0$ but $c, d$ are nonzero, then mod $5$ gives $a \equiv 0 \pmod 3$ and mod $7$ gives $a \equiv 0 \pmod 4$ so $a \equiv 0 \pmod {12}$, but mod $13$ means $5^c7^d \equiv 0 \pmod {13}$, which is false, so no solutions. If $d = 0$ and $a, b, c$ are nonzero, if $a = 1$ then we get $(1, 1, 1, 0)$ as a solution. By mod $4$ we see $a \le 1$ so $a = 1$ so consider when $c \ge 2$. By mod $9$ we have $c \equiv 3 \pmod 6$ and by mod $7$ this means $5^c = -1 \pmod 7$ so we need $2\cdot 3^b \equiv 0 \pmod 7$ which is false, so no solutions. If $c = 0$ and $b$ is nonzero, mod $3$ gives no solutions. Finally we consider when $a, b, c, d$ are all nonzero. Mod $5$ gives $a \equiv b \pmod 4$, so in addition with mod $7$ gives $a \equiv b \equiv 2, 4, 0 \pmod 6$. This means $a, b$ are even so we let $2x = a, 2y = b$ for positive integers $x, y$. Using difference of squares, we get $5^c7^d = (2^x3^y-1)(2^x3^y+1)$ so either $2^x3^y-1 = 5^c$ and $2^x3^y+1 = 7^d$, or the other way around so we have two cases. If $5^c-2 = 7^d$, then mod $7$ gives $c \equiv 4 \pmod 6$ so mod $9$ gives $7^d \equiv 5^c-2 \equiv 2 \pmod 9$. However, there does not exist positive integer $d$ where $7^d \equiv 2 \pmod 9$, so there are no solutions for this case. If $7^d-2 = 5^c$, if $c = 1$ then $d = 1$ so we get a solution $(2, 2, 1, 1)$. Otherwise $c \ge 2$, so by mod $25$ we see that there $7^d \equiv 7, 49, 43, 1 \ne 2 \pmod {25}$ which gives no solutions. Therefore, the solutions are $(1, 0, 0, 0), (3, 0, 0, 1), (1, 1, 1, 0), (2, 2, 1, 1)$. $\blacksquare$

The cases where $abcd=0$ can be dealt with by simple mods and Zsigmondy. Henceforth assume $abcd \neq 0$. First Step. If $a \geq 3$, then notice the following: By taking mod $3$, $c$ must be odd; By taking mod $8$, $c$ must be even and $d$ odd. This is an obvious contradiction. Second Step. Now assume $a=2$. This is perhaps the hardest case of all. Similar analysis of mod $3$ and mod $8$ yields $c$ and $d$ are both odd. Mod $35$ yields $b$ is even. Thus we have $$5^c 7^d = (2 \cdot 3^{b/2} + 1)(2 \cdot 3^{b/2} - 1).$$This means that $\{5^c, 7^d\} = \{2 \cdot 3^{b/2} + 1, 2 \cdot 3^{b/2}- 1\}$. Now Zsigmondy works again (!). Third Step. Finally, assume $a=1$. The congruence collapses to mod $8$ and mod $5$ now. Collecting all the edge cases, the final solutions are $(1, 0, 0, 0), (3, 0, 0, 1), (1, 1, 1, 0), (2, 2, 1, 1)$.

for the case when $wxyz=0$ work?

If $a \ge 3$, then by $\pmod 8$ and $\pmod 3$ we receive a contradiction. Therefore $a < 3$. Notice that $a=0$ obviously doesn't work. If $a=1$ and $c, d > 0$, then we have $3^b = 18 \pmod{35}$ which has no solutions. Therefore one of $c,d$ has to be $0$ and these cases are easy to check. Now if $a = 2$, notice that by $\pmod{35}$, $b = 2 \pmod{12}$, so $b$ is even. Let $b = 2k$. $4 \cdot 3^b - 1 = (2 \cdot 3^k + 1)(2 \cdot 3^k - 1) = 5^c7^d$. Now since they are relatively prime to each other, each one is equal to either $5^c$ or $7^d$. Now, Zsigmondy's finishes.

Use Zsigmondy if any of a,b,c,d are 0 (for example $b=0$ means one of the primes in the set $\{5,7\}\mid5^c7^d=2^a-1$ is s.t. $5\nmid2^k-1\forall k<a\implies a=4$ and then solve, or $7\nmid2^k-1\forall k<a\implies a=3$ and solve). Henceforth assume all of them are positive, and call the assertion of taking mod x f(x), and taking mod 2 of x g(x). We have $\{f(8),f(3)\}\rightarrow\{(g(d)=1,g(c)=0),(g(c)=1)\}$, contradiction, so $a < 3$. $a=1\implies3^b = 18 \pmod{35}$ which has no solutions. $$a = 2\stackrel{f(35)}{\implies}b = 2 \pmod{12}\implies g(b) = 0,3^b4 - 1 = (3^{b'}2 + 1)(3^{b'}2 - 1) = 5^c7^d;$$since they are relatively prime to each other, each one is equal to either $5^c$ or $7^d$; Zsigmondy also finishes here. We conclude the answers are $(1, 0, 0, 0), (3, 0, 0, 1), (1, 1, 1, 0), (2, 2, 1, 1)$.

We can handle all of the cases where $abcd = 0$. $\newline$ If $d = 0$, then we have $2^{a}3^{b} = 5^c + 1$. By Zsigmondy, the only solution here is $(a, b, c, d) = (1, 1, 1, 0)$ and $(1, 0, 0, 0)$. $\newline$ If $c = 0$ then we have $2^{a}3^{b} = 7^d + 1$. Using Zsigmondy again, we have $(a, b, c, d) = (3, 0, 0, 1)$ and $(1, 0, 0, 0)$. $\newline$ If $b = 0$ then $2^a = 1 + 5^{c}7^{d}$. Then using Zsigmondy, we have $(3, 0, 0, 1)$. $\newline$ If $a = 0$ then $3^b = 1 + 5^{c}7^{d}$. There are no solutions here. $\newline$ Now, taking$\pmod{8}$ for $a \geq 3$ we get $5^{c}7^d \equiv -1\pmod{8}$. Then $c$ is even and $d$ is odd. Taking$\pmod{3}$, we get $5^{c}7^d = -1\pmod{3}$, if $c$ even and $d$ odd, then it is not true, so we have a contradiction. Therefore, $a < 3$. $\newline$ If $a = 1$, then by$\pmod{5}$ $23^b \equiv 1\pmod{5}$ Then $b \equiv 1\pmod{4}$. From our previous results with$\pmod{8}$, this is impossible, so $a$ must equal $2$. $\newline$. For $a = 2$, $(b, c, d) = (2, 1, 1)$ is clearly a solution. There aren't any other solutions for $(c, d) \leq 1$. Taking$\pmod{35}$, we have $4(3^b) \equiv 1\pmod{35}$, so $b = 0\pmod{2}$ Then by difference of squares we have $5^{c}7^d = (2 \cdot 3^{\frac{b}{2}} - 1)(2 \cdot 3^{\frac{b}{2}} + 1)$. $\newline$ Since the two terms on the RHS are relatively prime, one of them must equal to $5^c$ and the other $7^d$. For $5^c = 2 \cdot 3^{\frac{b}{2}} - 1$, $(a, b, c, d) = (2, 2, 1, 1)$ is clearly possible, and by Zsig there are no other solutions. For $5^c = 2 \cdot 3^{\frac{b}{2}} + 1$, we have $5^c - 1 = 2 \cdot 3^{\frac{b}{2}}$. Since $4$ is a factor of $5^c - 1$, this is impossible. $\newline$ So all of our solutions are $(2, 2, 1, 1)$, $(3, 0, 0, 1)$, $(1, 0, 0, 0)$, $(1, 1, 1, 0)$.

The answer is $(1,0,0,0)$, $(3,0,0,1)$, $(1,1,1,0)$, and $(2,2,1,1)$. If any of $a,b,c,d$ are zero, we can use Zsigmondy to show that the first three are the only solutions. For example, when $d=0$, we need $5^c+1$ to only have 2 and 3 as prime factors, and $c=1$ already has both 2 and 3, so by Zsigmondy, everything above $c=1$ has at least one other prime factor, so only $c=0$ and $c=1$ can possibly work. We can use a similar approach for $a=0$, $b=0$, and $c=0$. From now, assume $a,b,c,d\geq 1$. The core idea of this problem is repeatedly taking suitable moduli. Claim: $a\leq 2$. Suppose otherwise. Taking mod 8 reveals that $$5^c7^d\equiv 7\pmod{8}.$$This means that $c$ is even and $d$ is odd. However, this means that $5^c\equiv 1\pmod{3}$ and of course $7^d\equiv1\pmod 3$. However, since $2^a3^b$ is a multiple of 3, taking the original equation mod 3 gives $$0-1\equiv 1\pmod{3},$$contradiction. Case 1: $a=1$. Then, we have $$2\cdot 3^b-5^c\cdot 7^d=1.$$Taking mod 5, we have $3^b\equiv 3\pmod{5}$, which means that $b\equiv1\pmod{4}$. However, taking mod 7, we have that $3^b\equiv 4\pmod{7}$, so $b\equiv 4\pmod{6}$, but this contradicts with out previous conclusion. Therefore, no more solutions in this case. Case 2: $a=2$. Taking mod 5 gives $3^b\equiv 4\pmod{5}$. $b\equiv 2\pmod{4}$. In particular, $b$ is even. Thus, factor as a difference of squares $$(2\cdot 3^{b/2}-1)(2\cdot 3^{b/2}+1)=5^c\cdot 7^d.$$Note that each term on the left side can only have 5 and 7 as prime factors. However, they only differ by 2, so it is not possible for both of them to be divisible by 5 or both to be divisible by 7, the only possibility is that one of them is a power of 5 and one of them is a power of 7. Claim 2: The only way a power of 5 and a power of 7 differ by 2 is 5 and 7. Consider $5^n\pm 2=7^m$. If $n\geq 2$, then taking mod 25 yields that $$7^m\equiv \pm2\pmod{25},$$neither of which is possible, as the cycle for 7 is $1,7,-1,-7,1,\dots$ Thus, we must have $2\cdot 3^{b/2}-1=5,$ so $b=2$. This gives the solution $(2,2,1,1)$.

nice Case 1.$ abcd=0$ By Zsigmondy, we want only two primes in the $u^n\pm 1$ factorization, so we can do only few cases giving us the answer here is $$\boxed {(a,b,c,d)=(1,0,0,0), (3,0,0,1), (1,1,1,0)}$$Case 2. $a,b,c,d>0$ $\pmod 3$ gives $c$ odd $\pmod 8$ gives $a<3$ $\pmod 5$ gives $2^a\cdot 3^b\equiv 1$ so $4\mid a-b$ $\pmod 7$ gives $2^a3^b\equiv 1 \pmod 7$ and if $a=1$ then $b$ odd and $3^b\equiv 4\pmod 7$ impossible, so $a\neq 1$ This gives us $a=2$ and $b$ even, so $$\left(2\cdot 3^{\frac{b}{2}}-1\right)\left(2\cdot 3^{\frac{b}{2}}+1\right)=5^c\cdot 7^d$$Both factor are relatively primes and the first is $-1\pmod 3$ so $$2\cdot 3^{\frac{b}{2}}=5^c+1 \hspace{0.5cm}\text{and}\hspace{0.5cm} 2\cdot 3^{\frac{b}{2}}=7^d-1$$So we can do Zsigmondy again and $d=1$ so $b=a=2$ and $c=1$, so the answer is $$\boxed {(a,b,c,d)=(1,0,0,0), (3,0,0,1), (1,1,1,0),(2,2,1,1)}$$

The solution set is $\boxed{(a,b,c,d) = (3,0,0,1),(1,1,1,0),(2,2,1,1), (1,0,0,0)}$. If $a=0$, modulo $2$ yields an obvious contradiction. If $b=0$, we have \[2^a-1 = 5^c7^d.\] Realize that $2^{12}-1$ contains a factor of $5$ and $7$ in it, so $a \le 12$ by Zsigmondy. Manually testing each case yields the first solution. If $c=0$, we have \[7^d+1 = 2^a3^b.\] The LHS cannot be a multiple of $3$, so we just have \[7^d+1=2^a.\] Now, Zsigmondy shows that the only possible solutions satisfy $d<2$, which yields the first and last solutions. If $d=0$, we have \[5^c+1 = 2^a3^b.\] Zsigmondy shows that only $c=0$ or $c=1$ works, yielding the second and last solutions. From this point on, assume that $abcd \neq 0$ as we have exhausted all of the cases where $abcd=0$. For the sake of contradiction, suppose that $a \ge 3$: Modulo $3$ gives $2^c \equiv -1 \pmod{3}$, so $c$ is odd. Modulo $8$ gives \[5 \cdot (-1)^d = -1 \pmod{8}, \] which is impossible. Thus, $a <3$. Now, we will consider this equation through more mods: Modulo $5$ yields \[2^a(3)^b \equiv 2^a \left(\frac{6}{2} \right)^b \equiv 2^a \cdot 2^{-b} \equiv 2^{a-b} \equiv 1 \pmod{5}.\] It follows that $4 \mid a-b$. Modulo $7$ gives \[2^a3^b \equiv 1 \pmod{7}.\] If $a=1$, we have \[6(2^{a_0}3^{b_0})^2 \equiv 1 \pmod{7},\] for nonnegative integers $a_0$ and $b_0$ (note that $a$ and $b$ are the same parity). However, $6$ is not a quadratic residue modulo $7$, so there are no solutions in this case. This means that $a=2$. It follows that $b$ is even as well, so let $b=2e$. We have \[4 \cdot 3^{2e}-1 = 5^c7^d\]\[\implies (2 \cdot 3^e-1)(2 \cdot 3^e+1) = 5^c7^d.\] The two terms are relatively prime up to a power of $2$, so a modulo $3$ argument yields \begin{align*} 2 \cdot 3^e-1 = 5^c \\ 2 \cdot 3^e+1 = 7^d. \end{align*} Zsigmondy applies one last time and gives $c=d=e=1$, the final solution in our solution set.

We first handle the cases when at least one of $a$, $b$, $c$, $d$ are 0: Four 0's: No solutions. Three 0's: $(1,0,0,0)$. Two or one 0: Isolate (either) lone term and apply Zsigmondy to get the solutions $(3,0,0,1)$, $(1,1,1,0)$. Assume $\min(a,b,c,d) \ge 1$. Mod 35 gives $a$, $b$ even while mod 3, 4 gives us $c$, $d$ odd. Thus our equation can be rewritten as \[x^2-35y^2 = 1, \operatorname{rad}(x) = 6, \operatorname{rad}(y) = 35 \implies x+y \sqrt{35} = (6+\sqrt{35})^k\] using Pell. Expanding using Binomial Theorem and comparing integer and irrational parts, we find the only solution $(2,2,1,1)$, giving our four solutions \[\boxed{(1,0,0,0), (3,0,0,1), (1,1,1,0), (2,2,1,1).}\]

I claim that the only solutions are the only solutions are $(1,0,0,0)$, $(3,0,0,1)$, $(1,1,1,0)$, and $(2,2,1,1)$. First, note that if we rearrange the equation, we get that \[2^a3^b=5^c7^d+1,\]the RHS of which must be odd. Therefore, $a\geq1$. Now, we will take cases on $b$. 1. $b=0$. Writing this in, we have that \[2^a-1=5^c7^d.\]First, I claim that $c=0$. This is because if $c>0$, then we have that $5\mid 2^a-1$, which means that we must have $4\mid a$, since the order of $2$ mod $5$ is $4$. This implies that \[2^4-1\mid 2^a-1 \iff 15\mid 2^a-1 \iff 3\mid 2^a-1=5^c7^d,\]an obvious contradiction. Therefore, we must have that $c=0$. This gives us that \[2^a-1=7^d \iff 2^a=7^d+1.\]Clearly, $(a,d)=(1,0)$ and $(a,d)=(3,1)$ are solutions. However, if $d>1$, by Zsigmondy's, we have that there must exist a prime $p$ that divides $2^a=7^d+1$ but not $7^1+1=8$. This is a contradiction, since no prime other than $2$ can divide $2^a$. Therefore, in the entire case where $b=0$, our only solutions are $(a,b,c,d)=(1,0,0,0)$ and $(3,0,0,1)$. 2. $b>0$. First, I claim that $c$ must be odd. This is because we have that \[2^a3^b=5^c7^d+1 \iff 5^c7^d\equiv 2\mod 3 \iff 5^c\equiv 2\mod 3,\]which only occurs when $c$ is odd. Therefore, since $c$ is odd, we must have that $c\geq 1$. Additionally, this also means that $5^c \equiv 5 \mod 8$. Since $7^d$ is always either $1$ or $-1$ mod $8$, this implies that \[2^a3^b=5^c7^d+1\equiv \pm 5^c+1\equiv \pm 5+1 \mod 8,\]meaning that $2^a3^b$ is either $4$ or $6$ mod $8$. This means that $a=1$ or $a=2$. We now use this to take cases. 2a. $a=1$. Writing this in, we have that \[2(3^b)-1=5^c7^d \iff 2(3^b)=5^c7^d+1.\]First, I claim that $d$ must be $0$. Note that since $c$ is odd and therefore $\geq 1$, we have that \[5\mid 2(3^b)-1 \iff 3^b\equiv 3 \mod 5,\]which only occurs when $b$ is $1$ mod $4$. However, if $d>0$, we have that \[7\mid 2(3^b)-1 \iff 3^b\equiv 4 \mod 7,\]which only occurs when $b$ is $4$ mod $6$, which implies that $b$ must be even. However, from before, we had that $c$ must be $1$ mod $4$, implying that $c$ must be odd, a contradiction. Therefore $d$ must be $0$. Now, writing this in, we get that \[2(3^b)-1=5^c \iff 2(3^b)=5^c+1.\]Clearly, $(b,c)=(1,1)$ is a solution. However, if $c>1$, note that by Zsigmondy's, we have that there must exist a prime $p$ that divides $2(3^b)=5^c+1$ but not $5^1+1=6$. This is a contradiction, since no primes other than $2$ or $3$ can divide $2(3^b)$. Therefore, since we must also have that $b>0$, there exist no other solutions in this case other than $(1,1,1,0)$. 2b. $a=2$. Again writing this in, we have that \[4(3^b)-1=5^c7^d.\]Now, note that since $c$ is odd and therefore $\geq 1$, we have that \[5\mid 4(3^b)-1 \iff 4(3^b)\equiv 1 \mod 5 \iff 3^b \equiv 4 \mod 5,\]which only occurs when $b$ is $2$ mod $4$. Therefore $b$ must be even. Let $b=2k$. Now by difference of squares, we have that \[4(3^b)-1=5^c7^d\iff 2(3^k)-1 \mid 5^c7^d \iff 2(3^k)-1=5^m7^n,\]for some non-negative integers $m$ and $n$. However, this is the exact same equation as (2a), which gave us that $k$ must be $1$. Using this, we get that \[5^c7^d=4(3^{2k})-1=4(3^2)-1,\]which gives us our only solution in this case of $(a,b,c,d)=(2,2,1,1)$. Finally, we have exhausted all cases, giving us that the only solutions are $(1,0,0,0)$, $(3,0,0,1)$, $(1,1,1,0)$, and $(2,2,1,1)$, finishing the problem.

Solved with david_kim_0202 pretty fun. Too late to change now, $x,y,z,w$ are respectively $a,b,c,d$ in this solution. Really long and tiring. We claim that the only quadruples $(a,b,c,d)$ of solutions are $(1,0,0,0) , (3,0,0,1)$ $(1,1,1,0)$ and $(2,2,1,1)$. It is easy to see that these quadruples indeed satisfies the given equation. Now we shall show that these are the only ones. We first do case work on the case where one of $a,b,c,d=0$. When $a=0$, we have that \[3^b=1+5^c7^d\]where it is clear that the right hand side is always even, while the left hand side never is. Thus, $a>0$. Now, when $b=0$, we similarly have \[2^a-1=5^c7^d\]But, by https://artofproblemsolving.com/community/c6h3039731p30667184 we have that $2^a-1$ has all prime factors under 8, if and only if $a \in \{1,2,3,4,6\}$. Of these, $a=1$ gives the solution set $(1,0,0,0)$, $a=2,4,6$ give no valid solution set since each of these make $2^a-1$ divisible by 3 and $a=3$ gives the solution set $(3,0,0,1)$. Further, when $c=0$ we have \[7^d+1=2^a3^b\]But, considering $\pmod{3}$ we see that $2^a3^b \equiv 7^d+1\equiv 1^d+1\equiv 2 \pmod{3}$, so we must have $b=0$ in order for the right hand side to not be divisible by 3, which is a case we have already checked. Finally, when $d=0$, we have that \[5^c+1=2^a3^b\]But, by Zsigmondy's Theorem, $5^c+1$ has a primitive prime divisor (a prime $p\nmid 5^1+1=6$) for all $c\geq 2$, which implies that we must have $c=1$, which gives the solution set $(1,1,1,0)$ and we are done. Now, it what proceed we assume that $a,b,c,d >0$. Then, note that considering the given equation $\pmod{5}$ we have \[1\equiv 2^a3^b - 5^c7^d \equiv 2^a3^b\]We know that $2^a \equiv 2 , -1 , -2 , 1 \pmod{5}$ when $a\equiv 1,2,3,0 \pmod{4}$ respectively. Further, $3^b \equiv -2,-1,2,1 \pmod{5}$ when $b\equiv 1,2,3,0 \pmod{4}$ respectively. Thus, by checking cases we can see that $2^a3^b \equiv 1\pmod{5}$ if and only if $a\equiv b \pmod{4}$. Then, in particular $a$ and $b$ are of the same parity. In the case where $a\equiv b\equiv 0 \pmod{2}$, we can let $a=2r$ and $b=2s$ for positive integers $r,s$. Thus, we obtain, \begin{align*} 2^{2r}3^{2s} - 5^c7^d &1\\ 2^{2r}3^{2s} - 1 &= 5^c7^d\\ (2^r3^s-1)(2^r3^s+1) &= 5^c7^d \end{align*}Further, it is clear that the $\gcd$ of the two given factors is at most 2, so we must have either $2^r3^s-1=5^c$ or $2^r3^s+1=5^c$. We will investigate these cases separately. When $2^r3^s-1=5^c$ we have \[5^c+1=2^r3^s\]But, by Zsigmondy's Theorem, $5^c+1$ has a primitive prime divisor (a prime $p\nmid 5^1+1=6$) for all $c\geq 2$, which implies that we must have $c=1$ which gives $r=s=1$ and the solution set $(2,2,1,1)$. Then, when $2^r3^s+1=5^c$ we have \[5^c-1=2^r3^s\]But, by Zsigmondy's Theorem, $5^c-1$ has a primitive factor (a prime $p\nmid 5^2-1=24$) for all $c>2$, which implies that $c\in \{1,2\}$ of which none work since $c=1$ implies $s=0$ and $c=2$ gives $r=3$ and $s=1$ which does not have a suitable value for $d$. Now, looking at this equation $\pmod{3}$, we have that \[1=2^a3^b - 5^c7^d \equiv -(-1)^c(1)^d = -(-1)^c \pmod{3}\]which requires $c \equiv 1 \pmod{2}$. Further, if $a\geq 3$, looking at the equation $\pmod{4}$ gives \[1=2^a3^b-5^c7^d \equiv - (1)^c(-1)^d \equiv - (-1)^d\]which required $d \equiv 1 \pmod{2}$. Further, looking at the equation again $\pmod{8}$ we have \[1=2^a3^b-5^c7^d \equiv -5^c(-1)^d \equiv -(-3)(-1)=-3 \pmod{8}\]since both $c$ and $d$ are odd. But, this is a clear contradiction, so there exists no solutions where $a>2$. We are left to deal with the cases where $a=1$ and $a=2$. This is little bit messier. We do it in two cases. Case 1 : $a=1$, $c$ and $d$ are odd. It is not hard to see that $5 \mid 2\cdot 3^b-1$ if and only if $b \equiv 1 \pmod{4}$ and $7 \mid 2 \cdot 3^b -1$ if and only if $b \equiv 4 \pmod{6}$. But, this requires $b$ to be odd and even respectively. So, only one of these is possible, which implies one of $c$ or $d$ is 0, which we have already checked. Case 2 : $a=2$, $c$ and $d$ are odd. It is not hard to see that $5 \mid 4\cdot 3^b-1$ if and only if $b \equiv 2 \pmod{4}$ and $7 \mid 4 \cdot 3^b -1$ if and only if $b \equiv 3 \pmod{6}$. But, this requires $b$ to be even and odd respectively. So, only one of these is possible at a time, which implies one of $c$ or $d$ is 0, which we have already considered. Thus, none of these cases have any new solutions, so we have found all solutions to the given equation, and we are done.

We claim that the only solutions are $(1,0,0,0)$, $(1,1,1,0)$, $(3,0,0,1)$, and $(2,2,1,1)$, which all clearly work. We start by dealing with the cases when one of the variables is $0$. If $a=0$, it is clear there are no solutions for parity reasons. If $b=0$, we have $2^a-1=5^c7^d$. We test values up to $6$ to see that $a=1$ gives the solution $(1,0,0,0)$ and $a=3$ gives the solution $(3,0,0,1)$. Since $7\mid 2^3-1$ and $5\mid 2^4-1$, Zsigmondy tells us that for any $a>6$ there is a prime $p\neq 5,7$ such that $p\mid 2^a-1$, so there are no more solutions. If $c=0$, we have $2^a3^b=7^d+1$. The right side is always $2\pmod{3}$, so we must have $b=0$, which we have already covered. If $d=0$, we have $2^a3^b=5^c+1$, which works similarly to the $b=0$ case. We can see that $c=0$ again gives us the solution $(1,0,0,0)$, while $c=1$ gives us the solution $(1,1,1,0)$. Then Zsigmondy tells us that for any greater $c$ there is a prime $p\neq 2,3$ with $p\mid 2^c-1$, so again there are no more solutions. Now we assume all four variables are nonnegative. By considering the expression$\pmod{3}$ we can see that $c$ is odd. Using this, we can determine that $5^c7^d\pmod{8}$ can only be $3$ or $5\pmod{8}$, which forces $a<3$. Now we just have to check these two cases. If $a=1$, then we have $2(3)^b-1=5^c7^d$. Taking$\pmod{5}$ gives us $b\equiv 1\pmod{4}$, but taking$\pmod{7}$ we get that $b\equiv 4\pmod{6}$, so this case is impossible. If $a=2$, then we have $4(3)^b-1=5^c7^d$. Taking $\pmod{5}$ we get that $b$ is even, so we can factor the left side as $(2(3)^\frac b2-1)(2(3)^\frac b2+1)$. Since these two factors have a $\gcd$ of $1$ and differ by only $2$, one of them must be $5^c$ and the other must be $7^d$. Thus any solution in this case must have $7^d-5^c=\pm2$. If $7^d-5^c=2$, clearly $d=c=1$ works, giving the solution $(2,2,1,1)$. To show there are no larger solutions, we can rewrite as $7^d-7=5^c-5$. By taking $\pmod{5}$, we get that $d\equiv 1\pmod{4}$. This means that $7^4-1=2400$ must divide the left side, but the right side cannot be divisible by $25$, contradiction. If $7^d-5^c=-2$, we can similarly rewrite as $7^d+7=5^c+5$. By taking $\pmod{7}$ we get that $c\equiv 4\pmod{6}$. However, if $c$ is even the right side is divisible by $3$, while the left side must be $2\pmod{3}$, contradiction. We have checked all cases, so these four are the only possible solutions, and we are done.

One thing to keep in mind that using parity we get that x must be > 0 Case 1 Of x,y,z,w >=1 Then we can say that 6a-35b=1 hence a diophantine is formed Now for a0 and b0 we get 6 and 1 but if we use the general formula of diophantine then we would get all the other values in negative integral world which is not possible Therfore, we have to take 6 and 1 as a and b Now, what is a and b? a = 2^(x-1)*3^(y-1) = 6 = 2*3 It implies that x=y=2 Similarly, b = 5^(z-1)*7^(w-1) = 1 It implies that z=w=1 Therfore we get one pair as (2,2,1,1) Case 2:- Let y = 0 Then we get that 2a - 35b = 1 Again using diophantine we get that (x,y,z,w) can be (1,0,0,0) and (3,0,0,1) Case 3:- Let z = 0 We get 6a-7b = 1 as the diophantine which gives (1,0,0,0) and (3,0,0,1) as solutions Case 4:- Let w = 0 We get the diophantine as 6a-5b =1 And the solutions are (1,0,0,0) and (2,1,1,0) Hence the solutions are (2,1,1,0);(1,0,0,0);(3,0,0,1);(2,2,1,1)