Prove that $4bc+3b+3c+2$ is never a perfect square

Question

How can I show that for every pair of non-negative integers $(b,c)$, the expression

$$4bc + 3b + 3c + 2$$

is not a perfect square?

Answer 1

To add to the answer by Thomas, with some integer $x,$ $1 + x^2$ is never divisible by any (positive) prime $q \equiv 3 \pmod 4.$ The quick reason for this is that $-1$ is not a quadratic residue $\pmod q.$ Writing the Legendre symbol $(-1|q) = -1.$

I wrote a short proof of the general fact at Prime divisors of $k^2+(k+1)^2$

Answer 2

Hint:

Multiply by $4$ and you get $(4b+3)(4c+3)-1$.

If this is a square then $(4b+3)(4c+3)$ can be written as the sum of two squares.