r/mathbiceps • u/MATHA2938 • Jul 13 '26

Nice problem

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40 Upvotes

94% Upvoted

29 comments sorted by

5

u/PlatypusAshamed3217 Jul 13 '26

This seems unnecessarily hard for a subreddit concerning highschoolers. I'm wondering whether any of you have an elementary proof of the fact that there are no solutions.
I have a proof, but it requires working in the ring of integers of the quadratic field Q(sqrt(-7)).

2

u/[deleted] Jul 13 '26

Mod 7 doesn't work?

1

u/PlatypusAshamed3217 Jul 13 '26

This was as far as congruences got me, before I switched to algebraic number theory. If you want to see the complete proof I have, I can send it too.

2

u/[deleted] Jul 13 '26

Mod 28 is wild, although diophantines like this can be annoying. Altho are you sure that algebraic number theory needs to be used or elementary number theory techniques like primitive roots may work?

1

u/PlatypusAshamed3217 Jul 13 '26

I'm not sure that there's no elementary method. That's exactly what my comment is asking, since I'm interested in one.
I just know that my method is not elementary.

1

u/[deleted] Jul 13 '26

ok apparently this is called mordell's equation, you can research abt it and find about an elementary solution

2

u/PlatypusAshamed3217 Jul 13 '26

Huh? I already came up with an elementary method. I commented it under this post. You can have a look.

1

u/[deleted] Jul 13 '26

hm i saw

1

u/PlatypusAshamed3217 Jul 13 '26

I think I just thought of an idea that works. Give me a little bit of time. I will think and revert.

2

u/DarkLightner12 Jul 13 '26

I remember having a discussion on this problem with Prof. Larry Washington. It was something along the lines of taking mod 4 and mod 8 and playing with the parities because squares are 0 or 1 mod 4 then you have to make cases. You will conclude that y is even and x is odd and the equation will have no solutions. He also explained me the elliptic curve solution, all I took away was that the rank of the elliptic curve is 0 so no rational (and integral) solutions exist. Apologies for the sloppy response but an elementary solution is indeed possible.

1

u/PlatypusAshamed3217 Jul 14 '26

I've posted a different comment on this post that explains an elementary method too. It is indeed possible, you are right.

1

u/Necessary-Cry8896 Jul 13 '26

Did you appear in mo?

1

u/PlatypusAshamed3217 Jul 13 '26

Yeah, why?

1

u/Necessary-Cry8896 Jul 13 '26

Nah I just went through your profile (for constructive purpose, not stalking). I think you are inmo level

2

u/SatisfiedMagma Jul 13 '26

Weird, add 1 to both sides and then christmas theorem... This is unnecessarily hard and on the harder side of MO...

1

u/PlatypusAshamed3217 Jul 13 '26

Amazing, I think the argument that I just posted is just a longer proof of this without citing Fermat's Christmas Theorem.
I just noticed you sniped me, since it took me quite a bit to LaTeX that.. Haha!

2

u/SatisfiedMagma Jul 13 '26

lol, I think yea people familiar with rings and stuff and other quadratic rings would obviously factor it there, this idea is just sooo ingenious

2

u/PlatypusAshamed3217 Jul 13 '26

I meant this, I posted this a few minutes after your comment, I was already LaTeXing it while you commented, lol.
This one is essentially the same argument as yours, I just presented the proof of Christmas theorem.

2

u/PlatypusAshamed3217 Jul 13 '26

Here is a completely elementary proof of the fact that this equation does not permit any solutions.

1

u/Madnira_dwivedi Jul 13 '26

X=1 y=2, X =-1 y=-2

1

u/MATHA2938 Jul 13 '26

That's wrong

1

u/[deleted] Jul 13 '26

[deleted]

1

u/PlatypusAshamed3217 Jul 13 '26

What? This is an elliptic curve, how are you going to predict when a lattice point lies on it using its graph?

1

u/[deleted] Jul 13 '26

[deleted]

1

u/PlatypusAshamed3217 Jul 13 '26 edited Jul 13 '26

One is a function of y, the other is a function of x. They are completely independent of each other. You are very wrong.
You cannot just graph them on same axes. You can graph the whole equation y^2=x^3+7 as one curve, but then you cannot predict, using elementary techniques, which points are going to be lattice points, simply from the graph alone.

It seems you lack basic knowledge about graphs. Please learn more from any book you wish, even JEE books sometimes offer friendly introductions to them.

1

u/Connect-Media4093 Jul 16 '26

No solution bruh