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:
 * $a b + d^2 = \paren {a - d}^2$

Then:
 * $a b + \paren {d - 1}^2 < \paren {a - d}^2$

Suppose $a b + \paren {d - 1}^2$ were square.

Then either:
 * $a b + \paren {d - 1}^2 = \paren {a - d - 1}^2$

or:
 * $a b + \paren {d - 1}^2 < \paren {a - d - 1}^2$

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

Suppose:
 * $a b + \paren {d - 1}^2 = \paren {a - d - 1}^2$

Since $c = 2 d$ it follows that $c - 2 = 2 \paren {d - 1}$

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

But :
 * $a b + \paren {d - 1}^2 = \paren {a - d - 1}^2$

Therefore:
 * $\paren {a - 2} b + \paren {d - 1}^2 = a b + \paren {d - 1}^2$

Thus:
 * $\paren {a - 2} b = a b$

which is absurd.

Therefore $a b + \paren {d - 1}^2 \ne \paren {a - d - 1}^2$.

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

Let $h = 2 \paren {a - d - f}$.

It follows that:
 * $\paren {a - b - h} = 2 \paren {f - b}$

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

But :
 * $a b + \paren {d - 1}^2 = f^2$

Thus:
 * $\paren {a - h} b + \paren {f - b}^2 = a b + \paren {d - 1}^2$

which is absurd.

Therefore:
 * $a b + \paren {d - 1}^2 \not < \paren {a - d - 1}^2$

Both possibilities have been shown not to be possible.

Hence $a b + \paren {d - 1}^2$ is not square.

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