Preselection tests for the full-time training

Because $2$ and $47$ are relatively prime numbers, $47$ divides $3^x-2^y$ if and only if $47$ divides $2^{4x}(3^x-2^y)$. But $$2^{4x}(3^x-2^y)=48^x-2^{4x+y}\equiv 1-2^{4x+y}(\mathrm{mod}47).$$ Therefore $47$ divides $3^x-2^y$ if and only if $2^{4x+y}\equiv 1(\mathrm{mod}47)$.

On the other hand, we have $2^{23}\equiv 49^{23}\equiv 7^{46}\equiv 1(\mathrm{mod}47)$. We deduce that the prime number $23$ is the order of $2$ modulo $47$. This implies that $2^{4x+y}\equiv 1$ mod $47$ if and only if $23$ divides $4x+y$.