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Prove/disprove that $x^3 - x + 9 = 5 y^2 $ has integer solution. i.e) There doesn't exist $x,y$ in integers satisfying above equation.

I tried using congruence and get if solution exist then $x$ should be congruent to $2\pmod{5}$. But after that, I tried something but I haven't get any conclusion. Please give me a hint if the statement is true.

Cloud JR K
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  • No square of an integer can be $2 \pmod 5$. Is this what you mean? – Toby Mak Sep 26 '19 at 08:31
  • @metamorphy thanks for pointing out ! – Cloud JR K Sep 26 '19 at 08:38
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    It does have a rational solution, $x=\frac {4}{5},y=\frac{33}{25}$. –  Sep 26 '19 at 08:58
  • @S.Dolan thanks for remarks But question is about integers – Cloud JR K Sep 26 '19 at 09:02
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    Cloud JR, @S.Dolan's discovery is very important. It implies that modulo any integer not a multiple of five there are integer solutions (because $5$ and hence also $25$ have a modular inverse). This strongly suggests that we may need tools other than modular arithmetic to settle this. – Jyrki Lahtonen Sep 26 '19 at 09:05
  • Which brings me to request for the source of the quesion. Is there a reason to believe that tools lesser than those developed to find integer points on elliptic curves are sufficient here? – Jyrki Lahtonen Sep 26 '19 at 09:10
  • @Jyrki Lahtonen.@S.Dolan I haven't know this before , my apologies. – Cloud JR K Sep 26 '19 at 09:11
  • @jyrki lahtonten , It was in exercise of notes sum of two square. But I have no idea what sum of square have anything to do with this. Btw, this is not my homework as the due date ends a month before. I haven't solve it. So now I am trying but i think It won't work , that's why I ask here – Cloud JR K Sep 26 '19 at 09:14
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    Is your question actually $x^2-x+9=5y^2$? – Toby Mak Sep 26 '19 at 09:17
  • No it's x^3. Not squared, I checked it . – Cloud JR K Sep 26 '19 at 09:18
  • Can you attach the notes "sum of two squares" if it is a PDF? Else, take a photo of the problem, and of the pages coresponding to the chapter. – Sarvesh Ravichandran Iyer Sep 26 '19 at 09:46
  • https://drive.google.com/a/iisertvm.ac.in/file/d/13k5w7mboGYxpY4QgKxWmMq1hZ_MPPuPO/view?usp=drivesdk – Cloud JR K Sep 26 '19 at 12:47

2 Answers2

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The substitution $U=5x, V=25y$ turns this equation into that of an elliptic curve with the short Weierstrass form $$ U^3-25U+1125=V^2. $$ According to its LMFDB entry the integer points on this elliptic curve are $(U,V)=(4,\pm33)$

There are no solutions with $V$ a multiple of $25$, so the original equation has no integer solutions.

Jyrki Lahtonen
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More of a comment, but there are too many already. Having read all the comments, it seems to me a strong possibility that the equation whoever produced the pdf problem sheet intended to set was $$x^2-x+9=5y^2$$

Put another way, the OP insists his Q has $x^3$, but I strongly suspect the Q whoever set the problem wanted him to solve had $x^2$, so it might be worth giving a quick answer to that. @JyrkiLahtonen has already given an excellent answer to the Q as the OP has given it. I hope it has piqued his curiosity for delights to come ...

So completing the square: $$(x-1/2)^2+35/4=5y^2$$ Multiplying by 4 and rearranging: $$5(2y)^2-(2x-1)^2=35$$

Now $2x-1$ must be divisible by 5, so writing it as $5z$ and $2y$ as $w$ we get $$w^2-5z^2=7$$ which is a standard Pell equation. They are discussed endlessly on this site, and it is not hard to show that it has no solution in integers, and a fortiori no solution with $w$ even and $z$ odd).

almagest
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