I'm trying to solve the Diophantine equation $x^5-2y^2=1$.

Here's my progress so far. We can write the Diophantine equation as $$\frac{x-1}{2}\cdot(x^4+x^3+x^2+x+1)=y^2.$$ If $x\not\equiv1\pmod{5}$, then $\gcd(\frac{x-1}{2},x^4+x^3+x^2+x+1)=1$, so both $\frac{x-1}{2}$ and $x^4+x^3+x^2+x+1$ must be perfect squares (note: $x^4+x^3+x^2+x+1>0$). In particular, $4(x^4+x^3+x^2+x+1)$ is a perfect square. Comparison with $(2x^2+x)^2$ and $(2x^2+x+1)^2$ forces $-1\leq x\leq3$. This results in the solutions $(3,\pm11)$.

If $x\equiv1\pmod{5}$, then we can write the Diophantine equation as $$\frac{x-1}{10}\cdot\frac{x^4+x^3+x^2+x+1}{5}=\left(\frac{y}{5}\right)^2,$$ where $\gcd(\frac{x-1}{10},\frac{x^4+x^3+x^2+x+1}{5})=1$, so both $\frac{x-1}{10}$ and $\frac{x^4+x^3+x^2+x+1}{5}$ must be perfect squares. Thus, $$x=10a^2+1,$$ $$x^4+x^3+x^2+x+1=5b^2.$$ Unfortunately, this is where I get stuck. I can substitute the first equation into the second, giving $$10000a^8+5000a^6+1000a^4+100a^2+5=5b^2,$$ $$2000a^8+1000a^6+200a^4+20a^2+1=b^2,$$ $$2000a^8+1000a^6+200a^4+20a^2=(b-1)(b+1),$$ $$5a^2(100a^6+50a^4+10a^2+1)=\frac{b-1}{2}\cdot\frac{b+1}{2},$$ but this doesn't seem to be making progress, even with modular arithmetic considerations.

  • $\begingroup$ You missed a detail. If $(u, v)=1$ and $uv$ is a square, you may conclude that $u$ and $v$ each is a square or the negative of a square $\endgroup$
    – jjagmath
    Aug 6 '21 at 21:53
  • $\begingroup$ I am aware of that, but $x^4+x^3+x^2+x+1$ is always positive, so it's not a problem here. $\endgroup$ Aug 6 '21 at 21:59
  • 1
    $\begingroup$ Ok, also, there's a typo, $(2x^\color{red} 2+x)^2$ and $(2x^\color{red}2+x+1)^2$ $\endgroup$
    – jjagmath
    Aug 6 '21 at 22:01
  • $\begingroup$ Thanks, corrected. $\endgroup$ Aug 6 '21 at 22:11

Also not a super elementary argument, but the only non-trivial result I use here is that $K=\Bbb Q(\sqrt{-2})$ has class number $1$, i.e. $\Bbb Z[\sqrt{-2}]$ is a PID.
Let $x^5-2y^2=1$ with $x,y\in\Bbb Z$. Note that $x$ is odd. We can factor the equation as $$x^5=(1-\sqrt{-2}y)(1+\sqrt{-2}y).$$ Let $d$ be a gcd of $(1-\sqrt{-2}y),(1+\sqrt{-2}y)$ in $\Bbb Z[\sqrt{-2}]$. Note that $d\mid 2$ and $d\mid x^5$. As $x^5$ is odd this implies that $d\mid 1$, i.e. the elements are coprime, hence (as $\Bbb Z[\sqrt{-2}]$ is a UFD) there is a unit $\varepsilon\in \Bbb Z[\sqrt{-2}]^\times=\{-1,1\}$ and $z=a+b\sqrt{-2}\in\Bbb Z[\sqrt{-2}]$ such that $1+\sqrt{-2}y=\varepsilon z^5$. As $(-1)^5=-1$ we may assume that $\varepsilon=1$. Then expanding the fifth power and comparing coefficients we get: \begin{align*} 1&=a^5-20a^3b^2+20ab^4\\ y&=5a^4b-20a^2b^3+4b^5 \end{align*} The first equation implies $a=\pm1$ and for $a=-1$ we don't get any solutions and for $a=1$ we get $b^4-b^2=0$, i.e. $b=0$ or $b=\pm1$. These correspond to $y=0$ and $y=\pm11$. Hence the only possible solutions for the original equation are $(1,0),(3,-11),(3,11)$.

  • $\begingroup$ A little more elementary would be nice, but this is pretty good! $\endgroup$ Aug 7 '21 at 4:55
  • $\begingroup$ As far as I understand, in this case PID-ness follows from the fact that $\mathbb{Z}[\sqrt{-2}]$ is an Euclidean domain, that is, there is an analogue of the Euclid's algorhitm. $\endgroup$
    – richrow
    Aug 7 '21 at 7:18
  • $\begingroup$ There's also a nice geometric proof of PID in this case: Look at an ideal $I$, and a nonzero element $x\in I$ with $\lvert x\rvert$ minimal, and suppose that $(x)\subsetneq I$. The ideal $(x)$ forms a lattice in $\mathbb{C}$ built out of rectangles whose short side has length $\lvert x\rvert$ and whose long side has length $\lvert x\rvert\sqrt2$. Any point in such a rectangle is less than a distance of $\lvert x\rvert$ away from a vertex. Then you can translate an element of $I\setminus(x)$ to get a contradiction to the minimality of $x$. $\endgroup$ Aug 7 '21 at 20:51

You don't really say if you just want to know the solutions, or if you want a nice elementary argument for why the solutions are only $(3, \pm 11)$, if you just want a proven answer and not an elementary argument the following works, its a bit overkill but its easier than thinking if you know these methods already:

The equation $x^5 - 2y^2 = 1$ considered over the rationals defines a hyperelliptic curve, of genus 2. So there is a big hammer called Chabauty's method that often determines all rational points on such curves. Our curve is isomorphic via change of variables ($y\mapsto 4y,x\mapsto 2x$) to the curve $$y^2 = x^5 - \frac{1}{2^5}$$ or even to an integral model $$y^2 = -2x^6 + 2x.$$

The computer algebra system Magma can determine the rank of the Mordell-Weil group of the Jacobian of this curve (using the integral model above) to be 1 (and hence Chabauty's method applies), using a generator Magma can also run Chabauty's method automatically in this case, and provably find all rational points:

> R<x>:=PolynomialRing(Rationals());
> H:=HyperellipticCurve(x^5-1/(2^5));       
> HH,ma:=MinimalWeierstrassModel(H);  
> a,b,P:=RankBounds(Jacobian(HH):ReturnGenerators);
> a,b,P;
1 1 [ (x^2 - 1/3*x, 22/9*x, 2) ]
> pts:=Chabauty(P[1]);
> pts;
{ (1 : 22 : 3), (1 : -22 : 3), (0 : 0 : 1), (1 : 0 : 1) }
> [P : P in pts][1]@(ma^(-1));  
(3/2 : -11/4 : 1)
> [P : P in pts][3]@(ma^(-1));
(1 : 0 : 0)

From this list we see that translating back to the original equation/curve the only interesting rational solutions are those you found already.

If you only wanted integral solutions to begin with there should be less high-tech methods to do this!

  • $\begingroup$ Very nice. I have some familiarity with Chabauty-Coleman. What prime are you using here? (It's a little hard for me to tell from the code) That being said, I would certainly have a preference for a more elementary argument, if possible. $\endgroup$ Aug 6 '21 at 22:23
  • 1
    $\begingroup$ @ThomasBrowning Indeed the builtin magma command uses a combination of Chabauty and the Mordell-Weil sieve, and picks the primes it uses automatically, I think in this case its using 3 and 59, but that isn't specified in the code. $\endgroup$ Aug 7 '21 at 7:19

Here is an "elementary" proof. The given diophantine equation $x^5 = 1+2y^2$ admits the obvious solution $x=1, y=0$. Exclude this trivial solution and consider $a=x^5$ as an integral parameter which one wants to represent as the value of the quadratic form $t^2+2y^2$, with unknown integers $(t,y)$. Geometrically, the problem is equivalent to find the points of the sublattice $\mathbf Z^2$ of $\mathbf R^2$ which belong to the ellipse with equation $a= t^2+2y^2$. Since the lattice is discrete and the ellipse is compact, the set $S$ of wanted points is finite. If $S$ is not empty and $t=1$, symmetry w.r.t. the $t$-axis imposes that card $S=2$. If one follows this elementary approach, the only reason for the hypothesis $a=x^5$ seems to be the quick growth of the 5-th power, which allows to determine $S$ without too many trials.

NB : As usual, "elementary" methods often mask the power - and slickness - of more elaborate methods. The general process at work here is, as suggested by the answer given by @leoli1, the representation of a positive integer by a binary quadratic positive definite form.

  • $\begingroup$ For fixed $a=x^5$, this approach works. But in order to determine that there are no other solutions, you would need to check every possible $a=x^5$. In particular, there are infinitely many ellipses you have to check, and the set of such ellipses is not compact. So a priori, there could be infinitely many points. $\endgroup$ Aug 8 '21 at 20:16
  • $\begingroup$ For example, $x^2=1+2y^2$ has infinitely many solutions, and your argument does not distinguish between $x^2$ and $x^5$. $\endgroup$ Aug 8 '21 at 21:00
  • $\begingroup$ You're quite right. A complementary argument is necessary, for instance to compare the rates of growth of $x^5$ and 2$y^2$, i.e. of $x^4 +..+x+1$ and $2y+1$. But then the shortest way would be to get outside $\mathbf Z$, and one might as well work "non elementarily" as @leoli1. $\endgroup$ Aug 9 '21 at 9:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.