If $x \ge 5 \in \mathbb Z$, then $x$ has a square mate $y$ with $y < x$.

Question

A pair of positive integers are called square mates if their sum $x + y$ is a perfect square.(The concept of square mates is contrived just for this problem.)

There's a positive square integer $a$ such that $a^2 \le x < (a + 1)^2$.

I thought of doing $x = a^2$ and $y = (a + 1)^2 - a^2$, since $a^2 > (a + 1)^2 - a^2 = 2a + 1$ if $a^2 = 5$ at least, which doesnt seem correct. If it were to be correct, then $x + y = (a + 1)^2$.

How do I go about solving this problem?

Answer 1

Let $x \in \mathbb{Z}$. Then there exists an integer $a$ such that $a^2 \leq x < (a+1)^2$. If $x = a^2$ then let $y=0$ and we are done. Else let $y = (a+1)^2 -x$. Then

$$y < (a+1)^2 - a^2 = 2a+1$$

Since $y \in \mathbb{Z}$ we can say that

$$y \leq 2a \leq 2\sqrt{x} < x$$

With the last inequality coming from the condition that $x \geq 5$.