How does $x^4+y^4=z^2 \implies x^4+y^4=z^4$?

Question

Why is the statement "the following cannot be satisfied" for $x^4+y^4=z^2$ more strong than for $x^4+y^4=z^4?$

More specifically, how does $x^4+y^4=z^2 \implies x^4+y^4=z^4?$

This statement was found on page 4 of the following document.

If $x^4+y^4=z^2$ has no positive integer solutions, $x^4+y^4=z^4$ cannot have any positive integer solution. — Jack D'Aurizio, Jul 15 '16 at 15:59
@DonAntonio Hi, why does what you commented cause the implication? I don't follow. — Jack Pan, Jul 15 '16 at 17:26
@JackD'Aurizio Hi, this is exactly what I'm asking, is why your comment is true. — Jack Pan, Jul 15 '16 at 17:26
If you have a solution to $;x^2+y^2=z^2;$ , then you also have one to $;x^2+y^2=(z^2)^2=z^4;$ . — DonAntonio, Jul 15 '16 at 17:29

score 5 · Accepted Answer · answered Jul 15 '16 at 17:23

5

Your question is mistaken: you have not correctly understood what is written in the document that you refer to. Of course that if $(x,y,z)$ satisfies $x^4 + y^4 = z^2$, it does not follow that it also satisfies $x^4 + y^4 = z^4$ unless $z^2 = z^4$ (i.e. $z \in \{-1,0,1\}$), which is very restrictive and not what that author meant to say.

What is meant there is that if $(x,y,z)$ satisfies $x^4 + y^4 = z^4$, then there exist another triple $(X,Y,Z)$ satisfying $X^4 + Y^4 = Z^2$ (and that triple is precisely $(X,Y,Z) = (x,y,z^2)$).

answered Jul 15 '16 at 17:23

Alex M.

35,207

3

Dear @Majid, you deleted your own answer at 18:21:22, and the (correct) answers given by me and "Topological cat" got downvoted at 18:23:36 and, respectively, at 18:21:37 (the latter, remarkably, only 15 seconds after the deletion of your answer!). Remarkable coincidence! Do I report it to a moderator, or would you rather apologize? And, in general, revenge votes are not the solution; learning mathematics better is. It is not our fault that you have graduated in mathematics yet you make gymnasium-level mistakes. – Alex M. Jul 15 '16 at 18:34

score 4 · Answer 2 · answered Jul 15 '16 at 15:59

4

If it had a solution $(x,y,z)$, then $(x,y, z^2)$ is solution of the first one

answered Jul 15 '16 at 15:59

Topological cat

439

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You got downvoted (like me) by @Majid: he felt embarassed that he gave the wrong answer which got him 3 downvotes. Ridiculous. (See also my own comment under my answer). – Alex M. Jul 15 '16 at 18:37
@AlexM. he first suggested an edit to my answer, changing "of the first one" to "of the second one". It might be a little confusing to correctly say which implies the other, though the concept is there, such things can confuse anybody, and I suppose that everybody can relate to this. So it's an understandable mistake, he didn't have to make such a big deal out of it! – Topological cat Jul 15 '16 at 18:47
As a side note, you also suggested a modification to his answer (turning it into a correct one), but I have rejected it: if a user is active on MSE, it is more polite to signal his errors (unless they are minor) in comments (as I have done); significant edits of other people's answers, even if mathematically correct, are intrusive and aggressive because they change the meaning originally intended by their authors. Even a mistaken answer deserves some respect. Otherwise, welcome to MSE! – Alex M. Jul 15 '16 at 19:02

How does $x^4+y^4=z^2 \implies x^4+y^4=z^4$?

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