The question would, naturally, be very straight forward if there was a $2xy$ instead of a $4xy$. Then it would simply be a matter of doing:

$$ x^2+2xy+y^2=1\\ (x+y)^2=1\\ \sqrt{(x+y)^2}=\sqrt{1}\\ |x+y|=1 $$

Clearly, there are infinitely many pairs of integers that differ by one so there is an infinite number of integer solutions to $x^2+2xy+y^2=1$. Unfortunately, the same principle does not apply to $x^2+4xy+y^2=1$ where if I attempt to construct a similar proof all I can do is:

$$ x^2+4xy+y^2=1\\ (x+y)^2+2xy=1\\ 2xy=1-(x+y)^2\\ 2xy=(1+x+y)(1-x-y) $$

And, from there, I have no idea how to proceed to complete the proof that there are an infinite number of integer solutions. I'm wondering whether I'm approaching the question entirely in the wrong way or if I am simply missing something. A nudge in the right direction would be much appreciated!

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    $\begingroup$ oeis.org/A001353 Scroll down to the 2nd reference to Michael Somos $\endgroup$ – user940 Mar 8 '12 at 16:41

Hint $\ $ Consider $\,f(x) = x^2 + 4y\ x + y^2\!-\!1\,$ as a quadratic in $\,x,$ where $y$ is constant. By Vieta its roots $\,x,x'$ satisfy $\ x+x' = -4y.\,$ Thus if $\,x\,$ is a root then so too is $\,x' = -4y-x.$

This yields a reflection symmetry $\ \, (x,y) \mapsto (-4y-x,\,y)\,$ on the solution space. Composing

this with the reflection symmetry $\,(x,y)\mapsto (y,x)\,$ yields the map $\,(x,y)\mapsto (-4x-y,x)$

which, iterated starting at solution $(1,0),\,$ yields infinitely many solutions

$$ (1,0),\ (-4,1),\ (15,-4),\ (-56,15),\ (209,-56),\ (-780,209),\ (2911,-78),\ \ldots$$

Sequence $\, 0,1,4,15,56,209,\ldots$ satsifies the recursion $\,f_{n+2} = 4 f_{n+1} - f_{n}\,$ as is easily derived.

This can be transformed to the Pell equation $\ X^2\! - 3 Y^2 = 1\ $ and studied using standard results on Pell equations. See also the comments on this sequence at OEIS sequence A001353.

  • $\begingroup$ Pardon the silly question... what does "$x$ is a root of $x^2 + bx + c$" mean? Isn't that just saying that $x^2 + bx + c = 0$?? $\endgroup$ – The Chaz 2.0 Apr 21 '12 at 21:53
  • $\begingroup$ To the anonymous user suggesting edits: they are being rejected by other users before I have a chance to see them. If you have questions please post them in comments and/or a new question. $\endgroup$ – Bill Dubuque Feb 24 '14 at 5:48


Have you learnt about Pell's equation?

Try adding a multiple of $y^2$ (on both sides) to complete the square on the left.

  • $\begingroup$ Using Pell's equation you would get $(x + 2y)^2 -3y^2 = 1$. How would you then prove it after this? $\endgroup$ – Vishwa Iyer Jan 22 '14 at 17:03
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    $\begingroup$ @Vishwa: The equation $a^2 - 3b^2 = 1$ has infinitely many solutions. Now $y=b$, $x = a -2b$ and you are done, aren't you? $\endgroup$ – Aryabhata Jan 22 '14 at 21:16

Following Aryabhata, write it as $x^2+4xy+4y^2-3y^2=1=(x+2y)^2-3y^2$ Let us define $z=x+2y$, so this becomes $z^2-3y^2=1$. Clearly $z=1,y=0$ is a solution. Now if we have a solution $(z,y)$ we observe that $(z',y')=(2z+3y,z+2y)$ is also a solution, because $$\begin {align}z'^2-3y'^2&=(2z+3y)^2-3(z+2y)^2\\&=4z^2+12zy+9y^2-3z^2-12yz-12y^2\\&=z^2-3y^2\\&=1 \end{align}$$ Given any solution we can find a larger one, so there are infinitely many.

A useful resource is this page

  • $\begingroup$ thanks this really helped. $\endgroup$ – Vishwa Iyer Jan 22 '14 at 23:55
  • $\begingroup$ I'll award the bounty as soon as I can. $\endgroup$ – Vishwa Iyer Jan 23 '14 at 0:02
  • $\begingroup$ Sorry but how did you "observed" that $(2x+3y,z+2y)$ is also a solution? :) $\endgroup$ – Shobhit Feb 19 '15 at 14:05
  • $\begingroup$ @ILUA: It comes from noting that $z=2, y=1$ is a solution, which you can do by inspection or by studying the continued fraction for $\sqrt 3$. Then there is Brahmagupta's identity that shows you how to combine solutions $\endgroup$ – Ross Millikan Feb 19 '15 at 15:10

It's easy... Keep 'Y' constant and vary 'X'. For single value of 'Y' You will get two values of 'X', Go on varying 'Y' you will get different value of 'X' for each 'Y'.

  • $\begingroup$ How do you know that $x$ will be an integer (or even real) solution, for a fixed $y$? This seems to be a vast oversimplification of the problem. $\endgroup$ – user61527 Feb 6 '14 at 3:16

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