Solvability of the Diophantine equation $x^{2} - y^{2} = 4z^{n}$?

Question

It is known that for every integer $z$ there are integers $x, y$ such that $x^{2} - y^{2} = z^{3}.$ In fact, given an integer $z$, taking $x := z(z+1)/2$ and $y := z(z-1)/2$ suffices.

But how is the integer solvability of the equation $x^{2} - y^{2} = 4z^{n}$ when $n \geq 5$? By intuition it seems that this Diophantine equation is unsolvable.

score 6 · Accepted Answer · answered Apr 27 '15 at 12:22

6

Every integer number $n$ that is not of the form $4k+2$ can be expressed as the difference of two squares. For instance: $$ x= z^n+1, \quad y=z^n-1 $$ is always a solution of: $$ x^2-y^2 = 4z^n. $$

answered Apr 27 '15 at 12:22

Jack D'Aurizio

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Ah, yes. Thanks! – Yes Apr 27 '15 at 12:28
Is it also necessary? – Yes Apr 27 '15 at 12:31
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@Chou: It is a necessary condition because $x^2-y^2\not\equiv 2\pmod{4}$. It is a sufficient condition since $n=d_1\cdot d_2$ and $d_1\equiv d_2\pmod{2}$ grant that $x=\frac{d_1+d_2}{2}$ and $y=\frac{d_1-d_2}{2}$ fulfill $x^2-y^2=n$. – Jack D'Aurizio Apr 27 '15 at 12:36

Solvability of the Diophantine equation $x^{2} - y^{2} = 4z^{n}$?

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