The Diophantine equation $x^5-2y^2=1$

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.

• 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 Aug 6 '21 at 21:53
• I am aware of that, but $x^4+x^3+x^2+x+1$ is always positive, so it's not a problem here. Aug 6 '21 at 21:59
• Ok, also, there's a typo, $(2x^\color{red} 2+x)^2$ and $(2x^\color{red}2+x+1)^2$ Aug 6 '21 at 22:01
• Thanks, corrected. 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 $$(1,0),(3,-11),(3,11)$$.

• A little more elementary would be nice, but this is pretty good! Aug 7 '21 at 4:55
• 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. Aug 7 '21 at 7:18
• 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 $x$. 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);
> pts;
{ (1 : 22 : 3), (1 : -22 : 3), (0 : 0 : 1), (1 : 0 : 1) }
> [P : P in pts]@(ma^(-1));
(3/2 : -11/4 : 1)
> [P : P in pts]@(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!

• 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. Aug 6 '21 at 22:23
• @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. 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.

• 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. Aug 8 '21 at 20:16
• For example, $x^2=1+2y^2$ has infinitely many solutions, and your argument does not distinguish between $x^2$ and $x^5$. Aug 8 '21 at 21:00
• 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. Aug 9 '21 at 9:40