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On page 462 of my edition of "History of the theory of numbers, Volume II, Diophantine Analysis" by L. E. Dickson, Dickson credites Leonardo Pisano with the following Theorem:

"If three of the four numbers a, b, a - b and a + b are squares, the fourth is congruent."

The example $a = 16$, $b = 9$, $a + b = 25$, so 7 is congruent is given. While the statement is obvious if $a > b$ and $a, a - b$ and $a + b$ are squares, there need to be some restrictions though, as 8 and 50 are not congruent, while 0, 4 and 25 are squares (take a = b).

I am interested in this for the following reason: Considering certain matrices associated with parallelograms suggested, that if we consider an arbitrary primitive pythagorean triple $k^2 + l^2 = h^2$, both $k^2 + h^2$ and $l^2 + h^2$ are congruent numbers. This would indeed be true, if the above theorem could always be applied to this situation.

Unfortunately, I can neither read Italian nor Latin, so I cannot read the original or the commentary by Genocchi cited by Dickson.

Therefore I would like to have a reference to a precise general statement of this theorem with accompanying proof, whether a commented translation of the original source or a standard number theory text. Thank you!

thomashennecke
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There is an annotated english translation of the "liber quadratorum" by L. E. Sigler, published by Academic press in 1987, that we, that is three pupils of mine and me, found in the SLUB Dresden.

The contents are also described in an article by R. B. McClenon, "Leonardo of Pisa and his Liber Quadratorum" in Vol. 26 of the American Mathematical Monthly from 1919, pp. 1 - 8, which I then found in the Wikipedia article "The Book of Squares".

The information needed can indeed be inferred from there or from the question linked by Bill Dubuque. It is a direct consequence of Leonardo Pisanos parametric solution of the congruum problem, which can be given like this:

If $a$ and $b$ are positive integers and $a > b > 0$, then there exist positive integers $r$ and $s$ such that

$$ r^2 + 4ab(a-b)(a+b) = (a^2 + b^2)^2 $$ and $$ (a^2 + b^2)^2 + 4ab(a-b)(a+b) = s^2 .$$

But if three of the numbers $a$, $b$, $a - b$, $a + b$ are squares, say $a$, $b$, $a - b$ as in our case, we may divide by $4ab(a-b)$ to get three rational squares in arithmetic progression with difference $a+b$, hence $a+b$ is a congruent number by definition. This goes wrong if $a = b$, but cannot, if $a > b$, as in our case.

thomashennecke
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