Construction of Rational Straight Lines Commensurable in Square Only whose Square Differences Commensurable with Greater/Lemma 2

Proof
Let $a$ and $b$ be square numbers such that $a > b$.

Let $c = a - b$ be even.

Let $d = \dfrac c 2$

From Lemma 1:
 * $a b + d^2 = \left({a - d}\right)^2$

Then:
 * $a b + \left({d - 1}\right)^2 < \left({a - d}\right)^2$

Suppose $a b + \left({d - 1}\right)^2$ were square.

Then either:
 * $a b + \left({d - 1}\right)^2 = \left({a - d - 1}\right)^2$

or:
 * $a b + \left({d - 1}\right)^2 < \left({a - d - 1}\right)^2$

It cannot be greater, without being equal to \left({a - d}\right)^2 because they are consecutive numbers.

Suppose:
 * $a b + \left({d - 1}\right)^2 = \left({a - d - 1}\right)^2$

Since $c = 2 d$ it follows that $c - 2 = 2 \left({d - 1}\right)$

Therefore from Square of Sum less Square:
 * $\left({a - 2}\right) b + \left({d - 1}\right)^2 = \left({a - d - 1}\right)^2$

But by hypothesis:
 * $a b + \left({d - 1}\right)^2 = \left({a - d - 1}\right)^2$

Therefore:
 * $\left({a - 2}\right) b + \left({d - 1}\right)^2 = a b + \left({d - 1}\right)^2$

Thus:
 * $\left({a - 2}\right) b = a b$

which is absurd.

Therefore $a b + \left({d - 1}\right)^2 \ne \left({a - d - 1}\right)^2$.

Suppose $a b + \left({d - 1}\right)^2 = f^2$ for some natural number $f$.

Let $h = 2 \left({a - d - f}\right)$.

It follows that:
 * $\left({a - b - h}\right) = 2 \left({f - b}\right)$

So from Square of Sum less Square:
 * $\left({a - h}\right) b + \left({f - b}\right)^2 = f^2$

But by hypothesis:
 * $a b + \left({d - 1}\right)^2 = f^2$

Thus:
 * $\left({a - h}\right) b + \left({f - b}\right)^2 = a b + \left({d - 1}\right)^2$

which is absurd.

Therefore:
 * $a b + \left({d - 1}\right)^2 \not < \left({a - d - 1}\right)^2$

Both possibilities have been shown not to be possible.

Hence $a b + \left({d - 1}\right)^2$ is not square.

Thus we have two square numbers whose sum is not square, as required.