2025/013) Prove that for every $n \in N$ the following proposition holds: $7|3^n + n^ 3$ if and only if $7 |3^nn^ 3 + 1$

Now 7 cannot be multiple of 7 because in that case 7 cannot be factor of either of them

Because 7 is prime so using Fermat's little theorem we have  

$n^6-1=0$

or $n^6 \equiv 1 \pmod 7\cdots(1)$

Now let $7| 3^n + n^3$

So multiplying  by n^3 we get

 $7| 3^nn^3 + n^6$

Or  $7 | 3^n n^3+1$

Similaly we can prove the only if part