Prove $\sqrt{-3+44a} \in N$ can't be a square root of a square of a natural number

Question

Prove $\sqrt{-3+44a} \in N$ can't be a square root of a square of a natural number.

$$ \begin{equation*} \begin{cases} a \in N \\ d \in N \\ -3+44a = d^2 \end{cases} \end{equation*} $$

Answer 1

Since a square root of the square of a natural number (a perfect square) equal to the natural number itself, we need to prove that $\sqrt{-3+44a}$ cannot be a natural number. Thus, we need to prove that $44a-3$ cannot be a perfect square.

We can prove this by $44a-3 \equiv 8 \pmod{11}$. Since a perfect square modulo 11 equals to one of the set {0, 1, 4, 9, 5, 3}, $44a-3$ is not a perfect square. Therefore, $\sqrt{-3+44a}$ can't be a square root of a square of a natural number.