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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.

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  • $\begingroup$ 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
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    $\begingroup$ 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
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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 $(1,0),(3,-11),(3,11)$.

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  • $\begingroup$ 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
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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!

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  • $\begingroup$ 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
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    $\begingroup$ @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
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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.

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  • $\begingroup$ 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

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