0

My question originates from Rational Points on Elliptic Curves, (Silverman & Tate), though has little to do with elliptic curves.

In chapter $V$: Integer Points on Cubic Curves, section $3$ it says

For equations of degree two, we observe that $x^2 - y^2 = m$ has only finitely many solutions, whereas $x^2 - 2y^2 = m$ often has infinitely many solutions.

I am curious how adding a $2$ as a $y$-coefficient change the number of solutions?

My understanding: it makes sense that the gaps between squares become larger and larger, so many values cannot be obtained from subtracting two squares, but what would change about subtracting twice a square from a square instead?

I suppose my question comes down to a bit of analytic number theory: I am asking about the differences between the densities\distribution of

  1. The square numbers
  2. The square numbers and doubles of the square numbers

I will make it clear that my question has little to do with the solving of these equations (because there are plenty of other questions about this topic) and has much more to do with the distribution of the above sets.

Clyde Kertzer
  • 2,904
  • 2
    Note that $x^2-y^2$ factors, whereas $x^2-2y^2$ does not – J. W. Tanner Mar 16 '23 at 20:17
  • 1
    You changed it in the title. It is the other way around. The reason is "Pell". Of course $(x-y)(x+y)=1$ in the integers allows only two factorisations: $1=1\cdot 1$ or $1=(-1)(-1)$. So the statement is not "fairly deep", but elementary number theory. For Pell's equation see for example here. – Dietrich Burde Mar 16 '23 at 20:23
  • The world of integers is discrete and unpredictable (undecidable), generally speaking. For example, $x^2=1$ has integer solutions while $2x^2=1$ does not. $x+2y=1$ has infinitely many integer solutions while $2x+2y=1$ has none. – Apass.Jack Mar 16 '23 at 20:28
  • The equation $x^2-dy^2=1$ is a very special equation, namely the norm equation $1=N(z)=x^2-dy^2$ of the ring of integers in the quadratic number field $\Bbb Q(\sqrt{d})$. Here $d$ is square free, to avoid trivial cases, like $d=1$. – Dietrich Burde Mar 16 '23 at 20:32
  • I didn't downvote. And of course the case $x^2-2y^2=m$ is very much connected to the case $m=1$. Silverman wrote it for $m=1$. Now you have changed the question. Don't do this. – Dietrich Burde Mar 16 '23 at 20:44
  • 2
    I changed the title to match the question in the body. @DietrichBurde: Does the linked-to duplicate cover the cases $m \not= 1$? – Brian Tung Mar 16 '23 at 20:48
  • @BrianTung In principle yes, but you would like to add another duplicate for that - see for example here in the example $m=2$. Silverman (and this is the important case) asks it in the elliptic curve book for $m=1$. The OP had cited it correctly before but now changed it. – Dietrich Burde Mar 16 '23 at 20:54
  • But why did you write first $x^2-2y^2=1$ in the body of your post ? This is what I remember from Silverman. Is it not correct? – Dietrich Burde Mar 16 '23 at 20:59
  • Anyway, I am sorry if you are not satisfied. It may be that this duplicate is not the best possible, but certainly it has been solved here (the name is generalized Pell equation, $x^2-2y^2=m$). I can try to replace it by a better duplicate. – Dietrich Burde Mar 16 '23 at 21:02
  • 1
    Did you see this post? What works for $m=2$, works in general. So it comes back to the case $m=1$ essentially. – Dietrich Burde Mar 16 '23 at 21:03
  • 1
    I don't know exactly what your real question is, but the way it is phrased has Pell Equation deeply built into it. It is not really about the density of squares vs. density of twice squares. It is rather about how close a square can be to twice a square. Consider the following. If $x^2-y^2$ is a (small) constant, then $x$ and $y$ need to be pretty close to each other. But for $x^2-2y^2$ to be a small constant, then we need $2y^2$ to cancel out most of $x^2$, so $x\approx \sqrt2 y$. The extra solutions come from our ability to approximate $\sqrt2$ as the ratio $x/y$ very well. – Jyrki Lahtonen Mar 16 '23 at 21:49
  • @ClydeKertzer: It has to be reopened first. But I think I can cast the tipping vote. – Brian Tung Mar 16 '23 at 22:58

0 Answers0