Solve $$\begin{cases}x^2-y^2=15\\x^2-5xy+5y^2=1\end{cases}.$$ I thought that the second equation is homogeneous, but it isn't. Can you give me a hint?

  • $\begingroup$ You have $15=(x-y)(x+y)$; set $t=x-y$, then $15/t=x+y$ so you can solve for $x$ and $y$ in terms of $t$ and then put them in the second equation. $\endgroup$ – Angina Seng Sep 13 '20 at 21:11
  • $\begingroup$ @10th grade I solved your problem. Show your attempts and I'll show my solution. $\endgroup$ – Michael Rozenberg Sep 13 '20 at 21:11
  • $\begingroup$ Note that, abstractly, the 'resolution' of these two equations is going to lead to a quartic in one variable or the other; that quartic will be solvable but there are no guarantees that it will be at all clean. $\endgroup$ – Steven Stadnicki Sep 13 '20 at 21:18
  • $\begingroup$ @StevenStadnicki, it's possible to see in advance that the quartic will be a quadratic in the square of the chosen variable (but I agree the solution may well be messy). $\endgroup$ – Barry Cipra Sep 13 '20 at 21:37
  • $\begingroup$ @StevenStadnicki Let's say you take the resultant with respect to $x$. Because of the symmetry $(x,y) \to (-x,-y)$, $y$ is a root of the resultant iff $-y$ is. Thus the resultant is a quartic polynomial in $y$, but is an even function, i.e. it is a quadratic in $y^2$. $\endgroup$ – Robert Israel Sep 13 '20 at 21:37

If you want to use a homogeneous equation, scale your two given equations to match the constant terms and subtract:




Then $76t^2-75t+14=0, t=y/x=(75\pm37)/152\in\{1/4,14/19\}$ from the quadratic formula. Render $y$ as either of these ratios times $x$ in $x^2-y^2=15$ to extract solutions for $x$ and back-substitute for $y$.


Hint: If you subtract the first equation from the second, the result is linear in $x$, so you can solve for $x$ in terms of $y$.


two hyperbolas, again with plenty of integer points.

$$ x^2 - y^2 = 15 $$

$$ x^2 -5xy + 5 y^2 = 1 $$

enter image description here

  • 1
    $\begingroup$ How is this a hint for solving the system? $\endgroup$ – Ramanujan Sep 13 '20 at 23:10
  • $\begingroup$ @ViktorGlombik I am in favor of people drawing diagrams. The previous question by the same OP was two circles, which were tangent, both had several integer lattice points, and the only intersection was such a point.. I drew that one by hand and scanned it... Here we have two evident intersections. $\endgroup$ – Will Jagy Sep 13 '20 at 23:18

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