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.
Triplets based equation
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$\begingroup$
number-theory
elementary-number-theory
diophantine-equations
open-problem
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0Anyone who is a Number Theory guru here, please let me know if someone posts an open problem, how is it handled here (Any guidelines??) – 2012-03-24
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0I think questions about the working of the site itself are best answered at [meta] best! @KVRaman – 2012-03-24
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0@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. – 2012-03-24
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0@KannappanSampath I do get it now. You mean about the guidelines I can find answer on meta.math.se (Of course!) Thanks – 2012-03-24
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0Just added tag "open-problem" because someone(on meta math) posted an answer to my question about this tag. – 2012-03-27
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0@KannappanSampath Thanks for Mathematics Meta (I got an answer from someone that there is an open-problem tag) – 2012-03-27
2 Answers
7
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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0Just a nitpick. The problem is problem #16 on the pdf. – 2012-03-26
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0@K V Raman! Thank you so much for your information. – 2012-03-26
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0Oops, thanks @SivaramAmnbikasaran. (Such nitpicks are productive :) – 2012-03-26
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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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0! Wow! very interesting solutions. Thank you. – 2012-03-26
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0@PeterPhipps I have removed my comment. (Sorry I didn't realize you added that note later) – 2012-03-26