Congruent number problem- one elementary statement We call a positive integer $n$ a congruent number if there is a rational right triangle with area $n$.
Equivalently, $n$ is a congruent number if  there exists positive integers $x$ and $y$ such that 
$$x^2\pm n=y^2.$$
Somewhere, I came across a statement:
If we take $a$ and $b$  such that from the set $\{a,b,a+b,a-b\}$ $3$  of the elements are squares, then the $4$th element  is a congruent number.
I tried to prove it, but I can't find a way to do it. I think that the second definition is the one more suitable here. 
For example, I tried to suppose that $a=x^2$, $b=y^2$ and  $a+b=z^2$. Then $a-b=x^2-y^2$. I think this could be just an elementary exercise to find a relation from definition?
 A: I'll prove that if three of the four integers $\{\,a,b,a+b,a-b\,\}$ are squares, then the fourth is a congruent number. By the way, this result was first mentioned by Leonardo of Pisa, better known as Fibonacci. 
First, let's get a correct definition of congruent: a positive integer $n$ is said to be congruent if there is a rational number $x$ such that both of the numbers $x^2\pm n$ are squares of rational numbers. 
Next, a couple of lemmas we'll prove later: 
Lemma 1. The positive integer $n$ is congruent if and only if $nk^2$ is congruent for all positive integer values of $k$. 
Lemma 2. If $n$ is the area of a right triangle with integer sides then $n$ is congruent. 
Now for the proof of the main result. 
For any positive integers $a>b$, the right triangle with legs $2ab$ and $a^2-b^2$ has integer sides (the hypotenuse is $a^2+b^2$) and area $ab(a+b)(a-b)$, so $ab(a+b)(a-b)$ is congruent by Lemma 2. Since $ab(a+b)(a-b)$ is congruent, and three of the numbers $a$, $b$, $a+b$, $a-b$ are squares, Lemma 1 implies that the fourth of these numbers is congruent. 
Now let's go back and prove the lemmas. 
Proof of Lemma 1. If $n$ is congruent, then the numbers $x^2\pm n$ are both squares, so $k^2(x^2\pm n)=(kx)^2\pm nk^2$ are both squares, so $nk^2$ is congruent. And if $nk^2$ is congruent for all $k$, then it's congruent for $k=1$, so $n$ is congruent. 
Proof of Lemma 2. If $a,b,c$ are positive integers with $a^2+b^2=c^2$ (so $a,b$ are legs of a right triangle with hypotenuse $c$), then both numbers $c^2\pm2ab$ are squares ($c^2+2ab=a^2+2ab+b^2=(a+b)^2$, etc.), so $2ab$ is congruent, but $2ab$ is four times the area of the triangle, so the area is congruent. 
