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I have the following problem: find the integers solutions of equation $$ x^2+7=y^5 \;. $$

Could someone help with this, please?

My approach:

After applying usual idea considering this equation in ring $Z[\alpha]$, where $\alpha$ is a root of $x^2-x+3=0$, I obtained that $(x-\sqrt{-7})=(a+b \alpha)^5$, for some $a$, $b$ $\in$ $Z[\alpha]$. After equaling coefficient of $\sqrt{-7}$, I got: $$ 32=5a^4b-70a^2b-49b^5 \;, $$ and I don't know how to find such pairs.

Björn Friedrich
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user44636
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    see here https://www.cambridge.org/core/services/aop-cambridge-core/content/view/S0017089500031918 i hope this will help you – Dr. Sonnhard Graubner Jan 08 '17 at 13:51
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    Also, it appears that you have $x^2-25=y^5-32$, FWIW. – abiessu Jan 08 '17 at 13:56
  • you will get $$x\pm 5$$ and $$y=2$$ – Dr. Sonnhard Graubner Jan 08 '17 at 13:57
  • You appear to be assuming that $\mathbb Z[\sqrt {-7}]$ is a ufd, which is not the case. See, e.g., this – lulu Jan 08 '17 at 14:03
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    Yes, but I work in $Z[\alpha]$, not in $Z[\sqrt{-7}]$, where I have ufd. – user44636 Jan 08 '17 at 14:05
  • Well...Unless I have messed it up (definitely possible) your second equation has no solutions, so something seems awry. I deleted my post which shows (I think) that your second equation has no solutions. Can you see the deleted post? – lulu Jan 08 '17 at 14:09
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    In case you can't see the deleted solution, it is this: we see that $b,|,32$. Work $\pmod 5$ to see that $b=5k+2$ which implies $b=2,32$. Now work $\pmod 7$ to eliminate both. – lulu Jan 08 '17 at 14:13
  • nope, I can't, could you post , please. – user44636 Jan 08 '17 at 14:14
  • I'll undelete for now. – lulu Jan 08 '17 at 14:14
  • Thank you, it seems that there is a mistake in my approach, because there is a solution as mentioned above. – user44636 Jan 08 '17 at 14:21
  • Well, either that or there's a mistake in what I wrote. One suggestion: go through your calculation and find the pair $(a,b)$ attached to the solution $(5,2)$. If you can produce a viable $(a,b)$ then obviously I blundered. – lulu Jan 08 '17 at 14:23
  • I don't understand why nobody has noticed that $\alpha = \frac{1}{2}(1+ \sqrt{-11})$ yet, so that $\sqrt{-7} \notin Z[ \alpha ]$. Does this make sense? – Crostul Jan 08 '17 at 14:41
  • @Crostul Indeed. The logic of the question really appears based in $\mathbb Z[\sqrt {-7}]$. – lulu Jan 08 '17 at 14:42
  • @user44636 I am re-deleting my post. There is an error, in that I assume that $b>0$. If we allow $b<0$ then $b=-8$ works $\pmod 5$ and the resulting congruence $\pmod 7$ does have a solution. – lulu Jan 08 '17 at 16:16
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    @Crostul It seems likely to me that that's a typo. $x^2+x-2$ is the correct minimal polynomial of $\alpha=(-1+\sqrt{-7})/2$. (Correction: it should have been $x^2-x+2$, and $\alpha=(1+\sqrt{-7})/2$) – Jyrki Lahtonen Jan 08 '17 at 17:33
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    @lulu $\Bbb{Z}[\alpha]$ is the ring of integers of $\Bbb{Q}(\sqrt{-7})$, and it is a UFD. Oops. Sorry, this was already pointed out. – Jyrki Lahtonen Jan 08 '17 at 17:35
  • Yes. Correct, @peterag (noticed my error during a trip to the fridge:-) – Jyrki Lahtonen Jan 08 '17 at 17:45
  • Thanks to everyone for fixing mistakes, but what about solution? Any ideas? – user44636 Jan 08 '17 at 17:48
  • Interesting question. Note that $x$ must be odd, and that there are no solutions with $y\leq2\times10^7$ other that $(x,y)=(\pm5,2)$ and $(x,y)=(\pm181,8)$. Also SAGE returns only these four solutions, and no others. – Servaes Jan 07 '20 at 16:55

1 Answers1

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A hint has been already been given to rearrange the equation as, $$(x-5)(x+5)=(y-2)(y^4+2y^3+4y^2+8y+16)$$ and the solutions $(x,y)=(\pm 5,2)$ already found.

There is another solution, $(x,y)=(\pm 181,8)$, which I found with just a spreadsheet; sorry but I cannot justify this other than $176*186=6*5456$, as much of your maths is above my understanding

I’ve thrown together code and checked up to $y=602248$ but found no other solutions.

Old Peter
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