How to prove that $7p + 3^p -4$ is not a perfect square?

Question

How to prove that $7p + 3^p -4$ is not a perfect square?

I calculated: $\left(\frac{7p+3^p-4}{p}\right) = \left(\frac{-1}{p}\right)$. So if $p \equiv 3 \mod 4$, the result is $-1$. So in that case, $7p+3^p -4$ can't be a square. But what about the case $p \equiv 1 \mod 4$? Any hints? Thanks in advance.

Answer 1

If $p=4k+1$, then:

$$7p+3^p-4\equiv -1+(-1)\pmod{4}=-2\pmod{4}.$$

But perfect squares must be 0,1 mod 4.

Answer 2

The squares modulo $4$ are

$$(4k)^2 = 4(4k^2) \equiv 0 \pmod{4},$$

$$(4k+1)^2 = 4(4k^2 + 2k) + 1 \equiv 1 \pmod{4},$$

$$(4k + 2)^2 = 4(4k^2 + 4k + 1) \equiv 0 \pmod{4},$$

and

$$(4k + 3)^2 = 4(4k^2 + 6k + 2) + 1 \equiv 1 \pmod{4},$$

but if $p = 4 \ell + 1$, then $$7p + 3^p - 4\equiv (-1)(1) + (-1)^{4 \ell}(-1)^1 \pmod{4} \equiv -2 \pmod{4},$$ which is equivalent to neither $0$ nor $1$ modulo $4$, so $7p + 3^p - 4$ cannot be a square if $p = 4 \ell + 1$.