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How would one determine solutions to the following quadratic Diophantine equation in three variables:

$$x^2 + n^2y^2 \pm n^2y = z^2$$

where n is a known integer and $x$, $y$, and $z$ are unknown positive integers to be solved. Ideally there would be a parametric solution for $x$, $y$, and $z$.

[Note that the expression $y^2 + y$ must be an integer from the series {2, 6, 12, 20, 30, 42 ...} and so can be written as either $y^2 + y$ or $y^2 - y$ (e.g., 12 = $3^2 + 3$ and 12 = $4^2 - 4$). So I have written this as +/- in the equation above.]

Thanks,

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Considering this as a quadratic in $y$, it is sufficient (and necessary) that

$n^4 - 4n^2(x^2-z^2)$ is a perfect square.

i.e.

$n^2 - 4(x^2 - z^2)$ is a perfect square, say $q^2$.

Rewriting gives us

$(n-q)(n+q) = 4 (x^2 - z^2)$

If $n=2k$ is even, then $q$ needs to be even too (say $2m$) and

$(k-m)(k+m) = (x-z)(x+z)$

If $n=2k+1$ is odd, then $q$ needs to be odd too (say $2m+1$) and

$(k-m)(k+m+1) = (x-z)(x+z)$

Thus you can pick any $m$, and try to factor the left hand side above (choosing the right one depending on the parity of $n$) into two terms of the same parity.

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  • $\begingroup$ Not sure of a parametric solution. It sort of depends on $n$. $\endgroup$ – Aryabhata Mar 7 '12 at 15:08
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We will consider the more general equation:

$X^2+qY^2+qY=Z^2$

Then, if we use the solutions of Pell's equation: $p^2-(q+1)s^2=1$

Solutions can be written in this ideal:

$X=(-p^2+2ps+(q-1)s^2)L+qs^2$

$Y=2s(p-s)L+qs^2$

$Z=(p^2-2ps+(q+1)s^2)L+qps$

And more:

$X=(p^2+2ps-(q-1)s^2)L-p^2-2ps-s^2$

$Y=2s(p+s)L-p^2-2ps-s^2$

$Z=(p^2+2ps+(q+1)s^2)L-p^2-(q+2)ps-(q+1)s^2$

$L$ - integer and given us.

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For the case when the equation: $X^2+qY^2+qY=Z^2$

factor $q$ - is not a square, then solutions can yrazit using Pell's equation: $p^2-qs^2=1$

Then there is another solution:

$X=X$

$Y=2psX-p^2$

$Z=Xp^2-qps+qXs^2$

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This is a good problem as it requires to write the (Diophantine) parametric representations of the variables.

This method can be extended to n variables; one way of stating this result is to say there is an algorithm to construct a n variable quadratic system and if our equation fits this construction then the solutions are read off trivial.

I will explain this later.

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  • 1
    $\begingroup$ So now is later... $\endgroup$ – Thomas Sep 15 '12 at 21:34
  • $\begingroup$ Is it "later" yet? :) $\endgroup$ – Ricket Sep 28 '14 at 3:53

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