# The Diophantine equation, $x^2 - y^2 = z^3$

I was recently studying some Diophantine equations and the equation $x^2 - y^2 = z^3$ caught my eye.I knew that $x=(n+1)(n+2)/2$ , $y=n(n+1)/2$ , $z=(n+1)$ where $n \in \mathbb{N}$ gives a set of positive integer solutions.I wanted to obtain a different form of solution when suddenly the identity

$\{(m+t)^2 - (m-t)^2 \}(4mt)^2 = (4mt)^3$ came into my mind so I could produce $x=4mt(m+t)$ ,

$y=4mt|m-t|$ , $z=4mt$ where $m,t\in \mathbb{N}$ ; I would really like to know is this a (well) known solution? Is there any solution in other forms?

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Well,thanks for the link. –  Souvik Dey Sep 21 '12 at 9:46

Write $z = a^m b^n$, where $m,n \geq 1$. Let $x+y = a^k b^l$ and $x-y = a^{3m-k} b^{3n-l}$, where $1 \leq k <3m$ and $1 \leq l <3n$. This gives us $$x = \dfrac{a^kb^l + a^{3m-k} b^{3n-l}}2 \text{ and }y = \dfrac{a^kb^l - a^{3m-k} b^{3n-l}}2$$ You can mimic the same thing with $z = a_1^{m_1} a_2^{m_2} \cdots a_k^{m_k}$.