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Was it already proved that $x^4-y^4=2$ has no solutions over rationals?

Or for simpler sub-problems $x^2-y^4=2$ or $x^4-y^2=2$.

azerbajdzan
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  • $ x^2 - y^4 = 2 $ (and the other one) isn't exactly a subproblem. Your original problem is a subproblem of any of those, not the other way around. – Yaël Nov 10 '19 at 20:50
  • We know that if there is a solution $x=\frac ab$, then $a=1$, $b\mid y^4+2$ by RRT. – Rushabh Mehta Nov 10 '19 at 20:51
  • @Astaulphe: I depends on the perspective. If you are looking for a proof or for a disproof. – azerbajdzan Nov 10 '19 at 21:03
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    The curve $x^4-y^4=2$ is a double covering of the elliptic curve $x^2-y^4=2$. You can check that this has rank 0. If you keep track of all the maps involved, you should be able to answer. – Ferra Nov 11 '19 at 17:56
  • Can you explain what you mean just describing one of the maps? – azerbajdzan Nov 11 '19 at 21:01
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    If you compute the rank using a program like Sage or Magma, usually you have to put the curve into Weierstrass form. Thus you get a birational map $\varphi$ from the curve $x^2-y^4=2$ to your curve into Weierstrass form, call it $C$. Then you can see that $C$ has a unique finite rational point, which is 2-torsion. Now remembering what $\varphi$ is, you recover all rational points on $x^2-y^4=2$, which are finitely many. Let $S$ be the set of such points. Any rational point on $x^4-y^4=2$ maps to a point of $S$ via the map $(x,y)\mapsto (x^2,y)$. – Ferra Nov 12 '19 at 11:22
  • @Ferra: But to be able to put $x^2-y^4=2$ into Weierstrass form you first need nontrivial rational point. But as I can not find any and at the same time I can not prove it does not exist I can not put it in Weierstrass form. So therefore I can not use Sage or Magma to compute its rank. – azerbajdzan Nov 12 '19 at 15:09
  • Or is Sage able to compute the rank also of elliptic curve in quartic form? – azerbajdzan Nov 12 '19 at 15:11
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    You have 2 non-trivial rational points on that hyperelliptic curve: they're the points at infinity. Anyway, check this answer: https://math.stackexchange.com/questions/1591990/birational-equivalence-of-diophantine-equations-and-elliptic-curves – Ferra Nov 12 '19 at 15:25
  • But as far as I know you can not use point at infinity to transform quartic elliptic curve into Weierstrass. You need "normal" rational point. At least I have not seen such transformation in literature. – azerbajdzan Nov 12 '19 at 15:28
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    If you read the answer I quoted you and you try to do the math yourself, you'll see that the rational map $x=G/6H$ and $y=(18x^2-H)/18$ gives you a rational map to an elliptic curve in Weierstrass form. If you don't like that the map is an isomorphism, you can also extend the scalars to $\mathbb Q(\sqrt{2})$ and hope the rank stays the same. – Ferra Nov 12 '19 at 16:25
  • I did such transformations many times myself, but always using normal point not point at infinity. What is $G$ and $H$? Can you paste here also transformations used? To verify it... – azerbajdzan Nov 12 '19 at 16:30
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    they are the new variables. Please read the answer I quoted! – Ferra Nov 12 '19 at 16:39
  • Yes I read it. And I know the method used. The "Cases 2" when $\ne \alpha^2$. But it requires a rational point $(p,q) $. How the method would work if I only have point at infinity $(0,1,0)$? So $z=0$. And right at the beginning I get $0=a$ because $z=0$ cancel out all terms and how to continue then? Then we have to use $y=w/z^2$ - but $z=0$, so again nothing to compute. – azerbajdzan Nov 12 '19 at 16:45
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    No, it's case 1. Probably you're confusing the curves. Here we're talking about $x^2=y^4+2$, which switching variable names is $y^2=x^4+2$. It is pretty clear it's case 1. – Ferra Nov 12 '19 at 16:52
  • I can not follow the cases 1. Where the variable $H$ comes from and $G$? They appear there as a miracle without saying anything about them, what they are, whats their definitions... – azerbajdzan Nov 12 '19 at 17:25
  • @Ferra: Finally I managed to do the transformation. Thank you. I upvoted your comments. – azerbajdzan Nov 12 '19 at 17:45

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We have the following elementary proof there's no solution. Let's rewrite the equation $$ (x - y)(x + y)(x^2 + y^2) = x^4 - y^4 = 2z^4 $$ with $ x, y, z \in \mathbb Z $ and $ \gcd(x, y, z) = 1 $.

For all prime $ p \ne 2 $, we have $ 4 \mid 4v_p(z) = v_p(x - y) + v_p(x + y) + v_p(x^2 + y^2) $. As the $ \gcd $ of any two of $ x - y, x + y, x^2 + y^2 $ is at most $ 2 $ (well known result from cyclotomic polynomials, or basic $ \gcd $ manipulations), we also have that at most one of $ v_p(x - y), v_p(x + y), v_p(x^2 + y^2) $ is nonzero.

So $ 4 \mid v_p(x - y), v_p(x + y), v_p(x^2 + y^2) $, which implies that $ x - y, x + y, x^2 + y^2 $ are all three a $ 4 $-th power times a power of $ 2 $ (which we can assume to be one of $ 1, 2, 4, 8 $ by "absorbing" higher powers in the $ 4 $-th power).

If one of $ x $ or $ y $ is even, then the other is too and $ 16 \mid 2z^4 $ which means that $ 2 \mid z, \gcd(x, y, z) $. Absurd. So $ x $ and $ y $ are both odd and $ v_2(x^2 + y^2) = 1 $.

We'll use the fact that $ 2(x^2 + y^2) = (x - y)^2 + (x + y)^2 $. We distinguish cases depending on how the $ 2 $s are distributed among the three factors:

  • $ x - y = a^4, x + y = b^4, x^2 + y^2 = 2c^4 $ which gives $ (a^2)^4 + (b^2)^4 = (2c^2)^2 $.
  • $ x - y = 2a^4, x + y = 8b^4, x^2 + y^2 = 2c^4 $ which gives $ (a^2)^4 + (2b^2)^4 = (c^2)^2 $.
  • $ x - y = 4a^4, x + y = 4b^4 $ implies that $ x = 2a^4 + 2b^4 $ is even (as well as $ y $). Absurd.
  • $ x - y = 8a^4, x + y = 2b^4 $ is the same as the second case but replacing $ y $ by $ -y $.

But $ a^4 + b^4 = c^2 $ is known to have no solutions in integers. One can prove this using the general form of a Pythagorean triple to derive an infinite descent.

Yaël
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  • "we also have that all three of them are 4-th power up to some factors of 2" - How did you deduce they are 4-th powers? I do not think it is true. What if, say, $x-y=6$. – azerbajdzan Nov 11 '19 at 20:58
  • Any prime $ p \ne 2 $ is such that $ 4 \mid 4v_p(z) = v_p(2z^4) = v_p(x^4 - y^4) $ and we know for sure that $ p $ doesn't divide any two of $ x - y, x + y, x^2 + y^2 $. So $ 4 \mid v_p(x - y), v_p(x + y), v_p(x^2 + y^2) $ which is exactly what I claim: $ \dfrac {x - y} {2^{v_2(x - y)}}, \dfrac {x + y} {2^{v_2(x + y)}}, \dfrac {x^2 + y^2} {2^{v_2(x^2 + y^2)}} $ are all $ 4 $-th powers. – Yaël Nov 12 '19 at 06:39
  • Sorry I do not get it. If $x=11, y=5$ then $x-y=23,x+y=2^4,x^2+y^2=273$ How this fits your arguments? – azerbajdzan Nov 12 '19 at 15:36
  • I'm assuming $ x^4 - y^4 = 2z^4 $. In your example, $ \dfrac {x^4 - y^4} 2 = \dfrac {11^4 - 5^4} 2 = 2^5 \cdot 3 \cdot 73 $ isn't a $ 4 $-th power so it doesn't even meet the requirements of your question. – Yaël Nov 12 '19 at 16:15
  • But you excluded such cases right at the begging of your proof. What if some of such cases are 4-th power? Also you are saying: "We see that this implies $x$ and $y$ are both odd which in return implies that $z$ is even." This is evidently wrong. If $x$ and $y$ are odd does not imply that $z$ is even. – azerbajdzan Nov 12 '19 at 16:27
  • Are you familiar with the concept of a proof by contradiction? We assume some assertion, then prove it leads to some absurdity to conclude that our assertion was wrong from the beginning. I assume that $ x^4 − y^4 = 2z^4 $ then show that it leads to another equation known for having no solution. Absurd. $ x, y, z $ such that $ x^4 − y^4 = 2z^4 $ didn't exist from the beginning. Or is it that you lost the logical progression of my argument? I've expanded myself on why $ z $ is even in my original answer. – Yaël Nov 12 '19 at 17:57
  • Yes, I Iost logical progression of your arguments. Now as you edited your answer I see that $z$ must be even. I still do not get the point that the terms must be 4-th powers times power of 2, but I was more concentrating on Ferra's comments using elliptic curves. I will look at your proof tomorrow again. – azerbajdzan Nov 12 '19 at 21:05
  • I clarified my explanations again by changing the arguments order. Hope you understand it now. – Yaël Nov 15 '19 at 17:59
  • What is the definition of $v_p(x)$? Is it how many times prime p divides x? – azerbajdzan Nov 16 '19 at 18:16
  • Yes, exactly. $ v_p(x) $ is the largest $ k $ for which $ p^k \mid x $. – Yaël Nov 17 '19 at 14:07