Squares and Pairs

Let $a$ and $b$ be any pair of integers an even distance apart, with $a<b$ , and $a \cdot b = z$
Let $x$ be the arithmetic mean of these two numbers. We know that $x$ must be an integer since $a$ and $b$ are an even distance apart.
Let $y=|x-a|=|x-b|$
Therefore, $a=x-y$ , and $b=x+y$
Plugging these into the first equation, we get $(x-y)(x+y)=z$
This is a difference of squares; we then know that $x^2-y^2=z$
Finally, that means $z+y^2=x^2$ , making the product of any two integers an even distance apart added to their distance to their arithmetic mean squared is always equal to their arithmetic mean squared.
Therefore, the answer must be Yes .

Let $a$ and $n$ be arbitrary integers. Then our pair of numbers will be $a$ and $a+2n$ .
The product of these integers are $a(a+2n)=a^2+2an$ .
By adding $n^2$ , we find that we have $a(a+2n)+n^2=a^2+2an+n^2=(a+n)^2$ .
So for any two integers a distance $2n$ apart, we know that adding $n^2$ will result in a square number.

There is an alternative and more concise proof, but that will be left as an exercise to the reader. :-)

Denote the two numbers as $k$ and $k + 2x$ . Their product is thus equal to $k\cdot(k+2x)$ . Note that

$k\cdot(k+2x)=(k+x-x)(k+x+x)=(k+x)^2 - x^2$

Clearly, the product $k\cdot(k+2x)$ can be written as a difference of two squares. Equivalently, as phrased in the problem, each product is some square number less than another square number .

For verification, here are the given items in the problem: $2\cdot4=2\cdot(2 + 2\cdot 1) = (2 + 1 - 1)(2 + 1 + 1) = (2+1)^2 - 1^2 = 3^2 - 1^2 = 8$ $20\cdot36=20\cdot(20 + 2\cdot 8) = (20 + 8 - 8)(20 + 8 + 8) = (20+8)^2 - 8^2 = 28^2 - 8^2 = 720$ $519\cdot93=93\cdot(93 + 2\cdot 213) = (93 + 213 - 213)(93 + 213 + 213) = (93+213)^2 - 213^2 = 306^2 - 213^2 = 48,267$