Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $p \ge 7$ be a prime number. Find the triples $(x, y, z)$ in $\mathbb{Z}$ such as $xyz$ is not equal to zero, $\gcd (x, y, z) = 1$ and $x^p + 2y^p = z^2$. I want triplets and proof/generalization. The reason for asking here, I am in position to construct equations and finding solutions by trail method. I am not in position to construct a proof or good generalizations. I hope, with your help, I can end.

share|cite|improve this question
Anyone who is a Number Theory guru here, please let me know if someone posts an open problem, how is it handled here (Any guidelines??) – Kirthi Raman Mar 24 '12 at 18:47
I think questions about the working of the site itself are best answered at Mathematics Meta best! @KVRaman – user21436 Mar 24 '12 at 18:51
@KVRaman Except for the "best" at the end, which was redundant, I have conveyed what I wanted to convey to the best of my ability. If you do not understand, it is better ignored. You may not understand it any later. – user21436 Mar 24 '12 at 19:21
@KannappanSampath I do get it now. You mean about the guidelines I can find answer on (Of course!) Thanks – Kirthi Raman Mar 24 '12 at 19:35
Just added tag "open-problem" because someone(on meta math) posted an answer to my question about this tag. – Kirthi Raman Mar 27 '12 at 0:00

There is a compiled list titled "SOME OPEN PROBLEMS ABOUT DIOPHANTINE EQUATIONS" and this problem happens to be listed as Problem#16 on page 3. Check it here

(Originally thought it was Problem #15, but I stand corrected it is indeed Problem#16).

share|cite|improve this answer
Just a nitpick. The problem is problem #16 on the pdf. – user17762 Mar 26 '12 at 5:17
@K V Raman! Thank you so much for your information. – gandhi Mar 26 '12 at 5:46
Oops, thanks @SivaramAmnbikasaran. (Such nitpicks are productive :) – Kirthi Raman Mar 26 '12 at 12:06

For $p=7$, let $a$ be a positive integer. Then
$$\begin{align*} (7a^2)^7 + 2(21.a^2)^7 &= 7^7a^{14} + 7^7a^{14}.2.3^7\\ &=7^7a^{14}(1+2.3^7)\\ &=7^7.a^{14}.4375\\ &=7^6a^{14}.175^2\\ &=(60025a^7)^2, \end{align*} $$ so there are an infinite number of triplets $(7a^2, 21a^2, 60025a^7).$

On re-reading the question, I see the gcd($x$, $y$, $z$)=1 constraint which kind of knobbles my answer. Ho hum.

share|cite|improve this answer
! Wow! very interesting solutions. Thank you. – gandhi Mar 26 '12 at 5:09
@PeterPhipps I have removed my comment. (Sorry I didn't realize you added that note later) – Kirthi Raman Mar 26 '12 at 21:32

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.