Proof: By Euclid

Let $A$ and $B$ be square numbers, and let $C$ and $D$ be their sides (respectively).
And let $A$ measure $B$ .
I say that $C$ also measures $D$ .

fig14e

For let $C$ make $E$ (by) multiplying $D$ .
Thus, $A$ , $E$ , $B$ are in continued proportion in the ratio of $C$ to $D$ [Prop. 8.11].
And since $A$ , $E$ , $B$ are in continued proportion, and $A$ measures $B$ , $A$ thus also measures $E$ [Prop. 8.7].
And as $A$ is to $E$ , so $C$ (is) to $D$ .
Thus, $C$ also measures $D$ [Def. 7.20] .
So, again, let $C$ measure $D$ .
I say that $A$ also measures $B$ .
For similarly, with the same construction, we can show that $A$ , $E$ , $B$ are in continued proportion in the ratio of $C$ to $D$ .
And since as $C$ is to $D$ , so $A$ (is) to $E$ , and $C$ measures $D$ , $A$ thus also measures $E$ [Def. 7.20] .
And $A$ , $E$ , $B$ are in continued proportion.
Thus, $A$ also measures $B$ .
Thus, if a square (number) measures a(nother) square (number) then the side (of the former) will also measure the side (of the latter).
And if the side (of a square number) measures the side (of another square number) then the (former) square (number) will also measure the (latter) square (number) .
(Which is) the very thing it was required to show.
∎

Thank you to the contributors under CC BY-SA 4.0!