For $p$ some prime, prove that $7p + 3^p - 4$ can never be a perfect square of a integer.
Let $p = 6a + b$ for some positive integers $a$ and $b$ where $0 \le b \le 5$. Thus since $p$ is a prime and $p=2$ or $p=3$ yield no solution, we can say that $p=6a+1$ or $p=6a+5$. Hence $p=6k\pm1 \quad ,k \in \mathbb{Z}$
For $p=6k+1$ we have: $$7\cdot (6k+1) + 3^{6k+1} - 4 = n^2$$
But the RHS is congruent to $\{0, 1, 2, 4\} \pmod7$ while the LHS is congruent to $6 \pmod 7$ for all $k \in \mathbb{Z}$
Thus we conclude that $p = 6k -1$. But then how can we prove that for $p=6k-1$ the above equality has no integer solutions?
Directly for $\;p=2\;$ : the expression is $\;14+9-4=19\;$ , not a square.
For general odd prime $\;p\;$, use $\;3^p=3\pmod 4\;$, so:
$$7p+3^p-4\equiv3(p+1)\pmod 4$$
But any squared EVEN integer (as our number. Check) has to be $\;0\pmod 4\;$ , so the above
happens only for $\;p=3\pmod 4\;$
But now we pass to modulo $\;p\;$ :
$$7p+3^p-4=3-4=-1\pmod p$$
and $\;-1\;$ is not a square when $\;p=3\pmod 4\;$