Prove that there is an integer $b$ greater than an integer $a$, such that $a + b$ is a perfect square.

Question

Let's represent the perfect square by $c^2$. Then:

$a + b = c^2$

$b = c^2 - a$.

Since, by definition, $c^2$ is an integer, $a$ is an integer and the subtraction of two integers yields an integer, $b$ exists and is an integer.

That is my solution, but it seems too simplistic. Could you please correct my proof or suggest a new one?

Answer 1

For given $a\in{\mathbb Z}$ the number $(a+1)^2$ is a square and sufficiently $>a$. Therefore let's try $$b:=(a+1)^2-a\in{\mathbb Z}\ .$$ Then $a+b$ is a square, and $$b-a=a^2+1>0\ ,$$ as required.