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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$\begingroup$ 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??) $\endgroup$ – Kirthi Raman Mar 24 '12 at 18:47
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$\begingroup$ I think questions about the working of the site itself are best answered at Mathematics Meta best! @KVRaman $\endgroup$ – user21436 Mar 24 '12 at 18:51
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$\begingroup$ @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. $\endgroup$ – user21436 Mar 24 '12 at 19:21
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$\begingroup$ @KannappanSampath I do get it now. You mean about the guidelines I can find answer on meta.math.se (Of course!) Thanks $\endgroup$ – Kirthi Raman Mar 24 '12 at 19:35
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$\begingroup$ Just added tag "open-problem" because someone(on meta math) posted an answer to my question about this tag. $\endgroup$ – Kirthi Raman Mar 27 '12 at 0:00
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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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$\begingroup$ Just a nitpick. The problem is problem #16 on the pdf. $\endgroup$ – user17762 Mar 26 '12 at 5:17
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$\begingroup$ @K V Raman! Thank you so much for your information. $\endgroup$ – gandhi Mar 26 '12 at 5:46
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$\begingroup$ Oops, thanks @SivaramAmnbikasaran. (Such nitpicks are productive :) $\endgroup$ – Kirthi Raman Mar 26 '12 at 12:06
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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.
answered Mar 25 '12 at 14:26
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$\begingroup$ @PeterPhipps I have removed my comment. (Sorry I didn't realize you added that note later) $\endgroup$ – Kirthi Raman Mar 26 '12 at 21:32
