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I want to find integers x,y,z, such that

k$z^2$ = $x^2$ - $y^2$

for a given integer k.

How do I write down the set of solutions? Preferably in parametric form.

For a given z, finding all the x and y is not difficult, I can just list the factors of k$z^2$ and use the factorisation (x+y)(x-y) = $x^2$ - $y^2$ to solve for x and y. So in principle, I can find all the solutions for z = 1,2,3,... There must be a clever way.

Noppawee Apichonpongpan
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  • Parametrizing rational solutions is not hard. Integer may be more complicated. Rewriting as $x^2-kz^2=y^2$ shows that for fixed $y$ we need to know all about Pell-type equations. – André Nicolas Jan 31 '15 at 09:54

1 Answers1

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Put it in this form,

$$y^2+kz^2 = x^2$$

Given the basic identity,

$$(m^2-v^2)^2+(2mv)^2=(m^2+v^2)^2$$

Let $v = n\sqrt{k}$. So,

$$(m^2-kn^2)^2+k(2mn)^2=(m^2+kn^2)^2$$

where $k=1$ gives you the Pythagorean triples.

Tito Piezas III
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