Can you find a simple proof for this statement?

I thought more about the last question I added into the addendum of the Numberphile, Graph theory and Mathematica post

It can be succinctly stated as:

$(\forall m\in\mathbb{Z}, m\ge 19) (\exists p,q\in\mathbb{Z}, 1\le p,q<m, p\ne q)$ such that $\sqrt{p+m}\in\mathbb{Z}$ and $\sqrt{q+m}\in\mathbb{Z}$ .

In words:

For all integers m, greater than 19, there are two other distinct positive integers less than m such that the sum of each with m, when square rooted is an integer.

What is the shortest proof you can find for this statement?

How clear is this post?

About the Author: Jonathan Shock

I'm a lecturer at the University of Cape Town in the department of Mathematics and Applied Mathematics. I teach mathematics both at undergraduate and at honours levels and my research interests lie in the intersection of applied mathematics and many other areas of science, from biology and neuroscience to fundamental particle physics and psychology.

2 Comments

Jonathan Rayner January 19, 2018 at 7:59 am - Reply

I’ll give it a shot:

To cheat a bit and simplify the argument, assume that we have verified the cases up to and including $m=23$ . By direct calculation

$\sqrt{m}(\sqrt{2} -1) > 2.02, \hspace{2mm} \forall m > 23$

(this is true for $m = 24$ and so true for all $m > 24$ ). So the interval $[\sqrt{m}, \sqrt{2m}]$ is of length >2, which means that that it contains two distinct natural numbers $a,b$ . Now:

$\sqrt{m} < a,b < \sqrt{2m}$
$\implies m < a^2, b^2 < 2m$
$\implies 0 < p, q < m$

where $p = a^2 - m$ , $q = b^2 - m$ . Then $p$ and $q$ satisfy the requirements of the question.
Jonathan Shock January 19, 2018 at 8:22 pm - Reply

Lovely

So, the next question is, what other sufficient conditions for non-traceability might exist?