# Triplets based equation

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.

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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 meta.math.se (Of course!) Thanks –  Kirthi Raman Mar 24 '12 at 19:35

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).

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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.

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! 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