An arithmetic sequence with no squares

Question

While playing with another problem, I found out that $a_n=4n+3;\;n\in\mathbb{N}$ contains no squares.

I tried to prove it in this way

$4n+3$ is odd so we must find an integer $m$ such that $4n+3=(2m+1)^2$ that is $$4n+3=4m^2+4m+1$$ Solving for $n$ I get $$n=m(m+1)-\frac{1}{2}$$ which is impossible for integer $n$.

Is this proof correct?

Answer 1

Your proof is correct, however I would finish the final part by instead saying that

$$4n = 4m^2 + 4m - 2$$

is always a multiple of 4 on the left hand side, and the right hand side can never be a multiple of 4. Or simply $0 \equiv 2 \mod 4$, which is a contradiction.