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I am trying to solve this diophantine equation: $x^2 + yx + y^2 = z^2$

In other words, I am trying to find integers $x$ and $y$ such that $x^2 + yx + y^2$ is a perfect square.

So far, the only methods to solve quadratic diophantine equations I am familiar with are Pythagorean triples and Pell equations. The $yx$ term has a coefficient of one, so I can't complete the square to reduce it to a Pell equation somehow, and I am wondering if there are other methods to solve this kind of equation.

Some insight would be highly appreciated.

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Generally speaking, this equation has a lot of formulas for the solution. Because it is symmetrical.

Write the formula can someone come in handy. the equation:

$Y^2+aXY+X^2=Z^2$

Has a solution:

$X=as^2-2ps$

$Y=p^2-s^2$

$Z=p^2-aps+s^2$

more:

$X=(4a+3a^2)s^2-2(2+a)ps-p^2$

$Y=(a^3-8a-8)s^2+2(a^2-2)ps+ap^2$

$Z=(2a^3+a^2-8a-8)s^2+2(a^2-2)ps-p^2$

more:

$X=(a+4)p^2-2ps$

$Y=3p^2-4ps+s^2$

$Z=(2a+5)p^2-(a+4)ps+s^2$

more:

$X=8s^2-4ps$

$Y=p^2-(4-2a)ps+a(a-4)s^2$

$Z=-p^2+4ps+(a^2-8)s^2$

For the particular case: $Y^2+XY+X^2=Z^2$ You can draw more formulas.

$X=3s^2+2ps$

$Y=p^2+2ps$

$Z=p^2+3ps+3s^2$

more:

$X=3s^2+2ps-p^2$

$Y=p^2+2ps-3s^2$

$Z=p^2+3s^2$

In the equation: $X^2+aXY+bY^2=Z^2$ there is always a solution and one of them is quite simple.

$X=s^2-bp^2$

$Y=ap^2+2ps$

$Z=bp^2+aps+s^2$

$p,s$ - integers asked us.

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  • $\begingroup$ I was looking for something like this! Can all solutions be generated from each case? or should one use all of them together to generate all solutions? $\endgroup$ – Ashura Jul 7 '15 at 4:34
  • $\begingroup$ @Ashura formally, we need to use all the formulas. But it is enough to use only General. Some formulas are mapped to each other. $\endgroup$ – individ Jul 7 '15 at 4:38
  • $\begingroup$ Can you possibly direct me to some resource with more information about this? Just out of curiousity. Thanks! $\endgroup$ – Ashura Jul 7 '15 at 4:47
  • $\begingroup$ @Ashura these formulas are a special case of a more General formulas. For example of such. math.stackexchange.com/questions/738446/… math.stackexchange.com/questions/794510/… Better in my blog to see. Many decisions have collected. artofproblemsolving.com/community/c3046 $\endgroup$ – individ Jul 7 '15 at 4:52
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We can generate solutions by a procedure similar to the one used to generate Pythagorean triples. Namely, it is equivalent to find rational points on the ellipse $a^2 + ab + b^2 = 1$. We do this by choosing a rational slope $m$, taking the line through $(1,-1)$ with slope $m$, and finding the other intersection of this line with the ellipse. This procedure gives the solution $a=-\frac{m^2+2m}{m^2+m+1}$, $b=\frac{2m+1}{m^2+m+1}$. Now, if $m=\frac{p}{q}$, we get a solution $a=-\frac{p^2+2pq}{p^2+pq+q^2}$, $b=\frac{2pq+q^2}{p^2+pq+q^2}$. Therefore, $x=-(p^2+2pq)$, $y=2pq+q^2$, $z=p^2+pq+q^2$ gives a solution to the original equation; and every solution in integers is a rational multiple of one of these solutions (since in reverse, the line through $(1,-1)$ and $(\frac{x}{z}, \frac{y}{z})$ has rational slope).

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Let $z$ be a fixed integer. We have $x^2+y^2+xy=z^2$ is equivalent to $(x+y)^2-xy=z^2$ and $(x-y)^2+3xy=z^2.$ Therefore, $(x+y)^2=z^2+xy\geq 0$ and $(x-y)^2=z^2-3xy\geq 0.$ Then, we have $z^2+xy$ and $z^2-3xy$ are squares. So, $|x+y|=\sqrt{z^2+xy}$ and $|x-y|=\sqrt{z^2-3xy}.$ This implies that $x+y=\sqrt{z^2+xy}$ or $x+y=-\sqrt{z^2+xy}$ and $x-y=\sqrt{z^2-3xy}$ or $x-y=-\sqrt{z^2-3xy}.$ You use after that the basic tools of calculus to find $x$ and $y$ separatly.

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  • $\begingroup$ I like what you did there, but it doesn't look like I can use those formulae to systematically generate all solutions to the equation above. Unless I'm missing something. $\endgroup$ – Ashura Jul 7 '15 at 4:29

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