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

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?

• Why do you get to assume $x$ is a square? $x=6$ satisfies $x \geq 5$, $x \in \mathbb{Z}$, and $x$ has a square mate $3 < x$. Commented Jul 6, 2014 at 22:54
• When can $(a+1)^2/a^2$ ever be as big as $2$? Why does that matter? Commented Jul 6, 2014 at 22:55
• @EricTowers are you serious? Commented Oct 16, 2014 at 0:58

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 $x \geq 5$.
• I don't get how $y < (a + 1)^2 - a^2$ if $y = (a + 1)^2 - x$ and $x = a^2$. Can you elaborate on that? Commented Jul 6, 2014 at 23:13
• In that case I am assuming $x \not = a^2$, thus $x > a^2$. Commented Jul 6, 2014 at 23:16