<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] Since $x,y,z \in \mathbb{N}$ we get that $5|5^x7^y$, so we have: $3^z \equiv 4 \mod 5$. Since $ord_5(3) = 4$ and $3^2 \equiv 4 \pmod 5$ we get $\fbox{$z = 4n + 2$}$. Obviously $z$ is even so we have $\fbox{$z=2k$}$. Then we have:
$$5^x7^y = 3^{2k} - 4 = (3^k - 4)(3^k + 4)$$
Now let $d = (3^k - 4,3^k + 4)$, then $d$ divides their diference, so $d\mid 8$. But both numbers are obviously odd, hence $(3^k - 4,3^k + 4) = 1$. Since we have 2 different prime factors on the left side we have $4$ cases:
Case 1:
\begin{cases} 3^k - 4 = 5^x \\ 3^k + 4 = 7^y \end{cases} 
Working wrt modulo 5 and 7 respectively we get $k=4n + 2$ and $k=6n + 1$, an obvious contradiction.
Case 2:
\begin{cases} 3^k + 4 = 5^x \\ 3^k - 4 = 7^y \end{cases}
Working modulo $3$ we get: $5^x \equiv 2^x \equiv 1 \pmod 3 \implies \fbox{$x=2m$}$
So we get: $3^k = (5^m - 2)(5^m + 2)$. As previously we get that the two factors on the right side are comprime. Since $5^m + 2 > 1$ we get that the only possibilty is:
\begin{cases} 5^m - 2 = 1 \\ 5^m + 2 = 3^k \end{cases}
For the first equation we have: $5^m = 3$ a contradiction.
Case 3:
\begin{cases} 3^k + 4 = 1 \\ 3^k - 4 = 5^x7^y \end{cases} 
An obvious contradiction for the first equation.
Case 4:
\begin{cases} 3^k + 4 = 5^x7^y \\ 3^k - 4 = 1 \end{cases}
From the second equation we get $3^k = 5$ a contradiction, because $k \in \mathbb{N}$ 
Hence this equation doesn't have a solution in $\mathbb{N}$
</s>