if a>0 then we can write$\displaystyle a^3=b^2-c^2$$\displaystyle b,c\in\mathbb{Z}$

how can ı prove the truthness this equation

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- Nov 24th 2008, 08:01 AMsah_mata number proof or disproof question
if a>0 then we can write$\displaystyle a^3=b^2-c^2$$\displaystyle b,c\in\mathbb{Z}$

how can ı prove the truthness this equation - Nov 24th 2008, 10:37 AMclic-clac
Don't you see that $\displaystyle \forall a \in \mathbb{N}^{*},\ a^{3}=\left( \frac{a^{2}+a}{2}\right)^{2}-\left( \frac{a^{2}-a}{2} \right)^{2}$ ? :p

Actually, you can write $\displaystyle a^{3}=(b+c)(b-c)$, and assume that $\displaystyle b,c$ are non-negative integers: since $\displaystyle b^{2}-c^{2}=(-b)^{2}-(-c)^{2}$ that doesn't change the result.

So a solution would be:

$\displaystyle b+c=a^{2}\ ,\ b-c=a$

You can solve that system in $\displaystyle \mathbb{Q}$, and find $\displaystyle b=\frac{a^{2}+a}{2} ,\ c=\frac{a^{2}-a}{2}$.

But $\displaystyle a^{2}$ and $\displaystyle a$ have the same parity, so the solutions found belong to $\displaystyle \mathbb{N}$, and give us a proof.