Question

Find a positive integer n such that $7n^{25}-10$ is divisible by 83.

Answer 1

Since $7\times 37=259\equiv 10mod83$
We have to find a value of n such that
$7n^{25}\equiv 7\times 37$ mod $83$
This is equivalent to
$n^{25}\equiv 37\equiv 2^{20}$ mod $83$
By Fermat's theorem
$2^{82k}\equiv 1\mathrm{mod}83$ for all k. So it is enough, if we choose n such that
$n^{25}\equiv 2^{82k+20}mod83$
If $k=15,$ this will be satisfied if
$n^{25}\equiv 2^{1250}\mathrm{mod}83$ and so if $n=2^{50}$
This gives one value of n.