74th NMO Selection Tests for JBMO

For any natural number $k$, we have $5^{2k}=M_8+1$, $5^{2k+1}=M_8+5$, $7^{2k}=M_8+1$ and $7^{2k+1}=M_8+7$, therefore from $8\mid 7^a-5^b$ we deduce that $a$ and $b$ are even. Denote $a=2m$, $b=2n$, with $m$ and $n$ positive integers. Then $7^{2m}-5^{2n}=8p$, i.e. $(7^m-5^n)(7^m+5^n)=8p$, with $p$ a prime. If $p=2$, then $(7^m-5^n)(7^m+5^n)\ne 16$. If $p\ge 3$, from $7^m-5^n<7^m+5^n$, as $7^m-5^n$ and $7^m+5^n$ are even, we have the following situations: $$(1^{\circ })\{\begin{array}{l}{7}^m-5^n=4\\ 7^m+5^n=2p\end{array},(2^{\circ })\{\begin{array}{l}{7}^m-5^n=2\\ 7^m+5^n=4p\end{array}.$$ Case (1°): From $7^m=M_3+1$ and $5^n=M_3\pm 1$, it follows that $7^m-5^n=M_3$ if $n$ is even, and $7^m-5^n=M_3+2$ if $n$ is odd. Since $4=M_3+1$, there are no solutions in this case. Case (2°): By subtracting the equations, we find that $2p=7^m-1$. Since $3\mid 7^m-1$, it follows that $3\mid 2p$, thus $p=3$. We deduce that $m=n=1$, $p=3$, therefore the solution is $(a,b)=(2,2)$.