<s>[INST] Find all positive integer solutions $(x,y,z)$ that satisfy $5^x \cdot 7^y +4= 3^z$? This is another contest math-problem.
The only problem that I cannot find the way to tackle this problem. 
Can anybody try to provide the solution to solve this problem?
Thanks
 [/INST] Given; $5^x \cdot 7^y +4= 3^z$ 
we get, $(-1)^x +1 \equiv 0 \pmod3$ hence $x$ must be odd, again, $(-1)^y \equiv (-1)^z \pmod 4$ hence $y$ and $z$ have the same parity.  Again $-1 \equiv 3^z \pmod 5$ or $1 \equiv 3^{2z} \pmod 5$ hence by Euler-Fermat theorem $2|z$. Again since $y$ and $z$ are even we have, $5^x \equiv 4 \pmod 8$, since $x$ is odd, we have two cases; either $x=4k+1$ or $x=4k+3$, note that $5^{4k} \equiv 1 \pmod 8$, the two cases each yield $5^x \equiv 5 \pmod 8$ a contradiction, hence $x$ cannot be greater than $0$, i.e $x=0$. The given equation then becomes $7^{2m} +4= 3^{2n}$ or $(7^m)^2 +2^2= (3^n)^2$, a pythagora's equation with solution, $2=2uv$, $7^m=u^2-v^2$ and $3^n=u^2+v^2$ from which it is obvious that $m$ and $n$ do not exist, and hence, $y(=2m)$ and $z(=2n)$ do not exist
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