2026/038) Find all $n \in N$ so that 7 divides $5^n+1$-.

 Basically we need to find n such that $5^n = -1 \pmod 7$

Now we have as 7 is a prime number as per Fermat's Little Theorem $5^6 \equiv 1 \pmod 7$

So $5^{6k} \equiv 1 \pmod 7$

Now as $5^6 \equiv 1 \pmod 7$ so we need to check for power of 5 to a factor of 6 that is 1 or 2 or 3

$5^1,5^2$ do not satisfy and $5^3 \equiv -1 \pmod 7$ satisfies.

so $n \in 6k+3 $ for all $k \in \mathbb{N}$