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Let d be a positive integer.The question is to prove that there exists a right angled triangle with rational sides and area equal to $d$ if and only if there exists an Arithmetic Progression $x^2,y^2,z^2$ of squares of rational numbers whose common difference is $d$

I tried using Heron's formula to get a relation between squares of sides and area but i failed and couldnot proceed.Is this an instance of an already known result I am unaware of?Any ideas?Thanks.

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Let $a,b$ the legs and $c $ the hypothenuse of the right angled triangle with rational sides and area equal to a given integer $d =\dfrac {ab}{2}$.

Then $\dfrac {(a-b)^2}{4}$, $\dfrac {c^2}{4} $, and $\dfrac {(a+b)^2}{4}$ form an arithmetic progression of squares of rational numbers with common difference $\dfrac {ab}{2} $.

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Hint:

If $y^2- d = x^2, y^2, y^2+d = z^2$, then: $z-x $ and $z+x $ are the legs of the right-angled triangle, area $= \frac {z^2-x^2}{2} = d $ and rational hypotenuse $2y $ because, $2 (x^2+z^2)=4y^2$.

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    Thanks for your answer.But how would you account for $d$ to be a positive integer. – 2017-02-16