MathSolver94 wrote: Let $m$ and $n$ be positive integers such that $2^n+3^m$ is divisible by $5$. Prove that $2^m+3^n$ is divisible by $5$. By euler, looking mod $4$ gives $m,n\equiv 1,3$ or $m,n\equiv 2,4/4,2$ and table of rests is $3,4,2,1/2,4,3,1$ and easy to find it is true. litle other: if $m$ is odd, $n$ is odd or they are both even, so $2^n\equiv (-3)^m$ and so, gives $2^m+3^n\equiv 3^m+2^n$ if they are even and $2^m+3^n\equiv -(3^m+2^n)\pmod{5}$ if they are odd.