Solution to Problem in the First Post

Well, I thought it would've taken longer for me to solve the problem but after more effort on observation I succeeded. Here goes the solution:

Problem: Let \(p\) be a prime, prove that \(7p+3^p-4\) cannot be a perfect square

Solution:

When \(p=2, 7*2+3^2-4=19\) is not a perfect square.
When $p$ is an odd prime, suppose there exists $p$ such that  $7p+3^p-4$ is perfect equare. Since $7p+3^p-4=n^2$ is even, therefore $7p+3^p-4\equiv 7p-1\equiv 0\pmod 4\Rightarrow p\equiv 3\pmod 4$. Next, note that $n^2+1\equiv 7p+3^p-3\equiv 0\pmod p$ by Fermat's Little theorem. However, it's well known that numbers of the form $n^2+1$ only have prime divisors$\equiv 1\pmod 4$, thus we've reached a contradiction and established \(7p+3^p-4\) cannot be a perfect square. $\Box$

Note to Solution: Before finding this solution, I tried examining different modules, especially $\pmod 6$ with high hopes because it has proven to be effective on problems with primes.

Next Challenge: Find all positive integers $n$ and $p$ if $p$ is prime and \[ n^8 - p^5 = n^2+p^2 . \]