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I have a nice diophantine equation which I tried to solve since march but no solution. Tried modulo 11, tried to write it in some way to figure out a solution... I posted this a few months ago, but it was removed. The problem: Prove that the equation $x^2-y^{10}+z^5=6$ has no integer solutions( positive, negative). Maybe another idea if YOU have or a solution! I would appreciate it very much!

EDIT, Will Jagy: If $$ x \equiv 4,7; \; \; \; y \equiv 0; \; \; \; z \equiv 1,3,4,5,9 \pmod {11} $$ THEN $$x^2-y^{10}+z^5 \equiv 6 \pmod {11}$$ which you can just check. At the level of school mathematics, there are two tricks I know for this type of problem. Neither works, but other people have found dozens to hundreds of tricks that I don't know. I also know of published problems with no answer because the book screwed up. Apparently the answer will be published in late September.

Pedro
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  • It's a youth mathematics magazine from Romania:) –  Sep 20 '13 at 19:46
  • I had already checked the equation mod the primes from 2 to 10^6, all have solutions. –  Sep 20 '13 at 19:59
  • Oh...I'll try it again, I must have made a mistake. –  Sep 20 '13 at 20:02
  • Can you please post a complete solution? What if they are congruent to 0 mod 11? I didn't understand the answer with factorising from mister Lin :( –  Sep 20 '13 at 20:27
  • No, you had it correct, if $y$ is divisible by 11 it can work mod 11, because 5 is a quadratic residue. So, more work required. – Will Jagy Sep 20 '13 at 20:30
  • @user85046 My solution is wrong, hence removed. – Calvin Lin Sep 20 '13 at 20:33
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    Why was the problem deleted before? – Will Jagy Sep 20 '13 at 22:47
  • Because they said it was a homework...The question obtained a lot of negative votes and it was closed. I hope we will solve it soon. –  Sep 21 '13 at 07:12
  • The earlier post was http://math.stackexchange.com/questions/436282/prove-that-the-equation-x2-y10-z5-6-has-no-integer-solutions where several commenters were convinced that it sufficed to work modulo 11. – Gerry Myerson Sep 21 '13 at 12:52
  • I don't think so. Can you show me that it works? If yes, then I really want to see this! :) –  Sep 21 '13 at 14:49
  • Did someone find something? :) –  Sep 21 '13 at 16:14
  • So, when does the Romanian youth magazine post the answer(s)? – Will Jagy Sep 21 '13 at 20:11
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    Hello! I thought nobody will return to this question! Well...at the end of this month. –  Sep 21 '13 at 20:22
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    CFG, in that case post that as an answer here once it appears. Meanwhile, you would have a better experience at http://www.artofproblemsolving.com/ and/or https://www.awesomemath.org/ – Will Jagy Sep 21 '13 at 20:30
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    Avoid subjective titles, make them informative. – Pedro Sep 21 '13 at 20:32
  • Modulo $41$ and $61$ there are solutions to the congruence. Modular arguments seem hopeless. – Bart Michels Jun 14 '14 at 09:59
  • @WillJagy: Could approaches of the form $~(x^2-y^{2\cdot5})+z^5=6,~$ or $~x^2-(y^{5\cdot2}-z^5)=6,~$ or $~(x^2-1)-(y^{10}+4)+(z^5-1)=0,~$ etc. be of any help ? – Lucian Aug 23 '20 at 07:26
  • @Lucian I guess I will fiddle with it. An answer appeared in that Romanian magazine in 2013, so there are likely peculiar tricks (assuming problem correctly transcribed) – Will Jagy Aug 23 '20 at 15:03
  • @WillJagy: Do you by any chance remember the name of the publication, because I've checked a few, but couldn't find it anywhere, either proposed or solved. – Lucian Aug 23 '20 at 15:44
  • @Lucian all the OP said (in 2013) was "It's a youth mathematics magazine from Romania:)" Romania was an SSR for quite a while, might be in Russian. I see Gerry Myerson linked a previous version of this. One possibility https://www.ssmrmh.ro/category/proposed-problems/ – Will Jagy Aug 23 '20 at 16:24

1 Answers1

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Partial answer:

According to Fermat's little theorem, for any prime $p$ and any integer a, $a^p\equiv a \pmod p$.

Working mod 2: \begin{align} x^2-y^{10}+z^5 &\equiv 6 \pmod2\\ x-y+z &\equiv 0\pmod2\\ y &\equiv x+z\pmod 2 \end{align} Working mod 3: \begin{align} x^2-y^{10}+z^5 &\equiv 6 \pmod3\\ x^2-(y^3)^3y+z^3z^2 &\equiv 0 \pmod 3\\ z &\equiv y^2-x^2 \pmod3 \end{align} Now, any square is 0 or 1 mod 3. So if $3\nmid x$ and $3\nmid y$ then $3 \mid z$, or $3 \mid xyz$.

Suppose that $g$ is the greatest common divisor of $x$, $y$ and $z$. Then $g^2 \mid 6$, implies $g=1$.

Kieren MacMillan
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LeeNeverGup
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