Do you know the identity to solve this?

N $o$ t $i$ c $e$ t $h$ a $t$ $2500=50^2=4.5^4$ .

The identity to solve this is the SG identity.

If $n$ were even, it (the number) would never be a prime.

So, $n$ is odd. Let $n=2k+1$ .

Then, $2500^n+n^{2500}=2500.2500^{2k}+n^{2500}$ , which can further be written as:- $4.(5 \times 50^k)^4+(n^{625})^4$ , which can even further be written as $4a^4+b^4$ , which is factorisable.

$\bullet \quad b^4+4a^4 = (b^2+2ab+2a^2)(b^2-2ab+2a^2)$

We can not generalize this for all positive integers. As a simple counter example, for ANY $n$ that is a divisor of $2500$ this identity does not hold.

For example, let us check for $n=50$ .

$2500^{50}+50^{2500}=50(50^{50}+50^{2499})$ , meaning in this case the number we get from the expression is divisible by $50$ , therefore it is not a prime number.