Integer solution of $x^3 - x + 9 = 5 y^2$

Question

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.

No square of an integer can be $2 \pmod 5$. Is this what you mean? — Toby Mak, Sep 26 '19 at 08:31
It does have a rational solution, $x=\frac {4}{5},y=\frac{33}{25}$. — , Sep 26 '19 at 08:58
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 [email protected] 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
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

Jyrki Lahtonen · Accepted Answer · 2019-09-26T10:00:00.027

5

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.

edited Sep 26 '19 at 10:00

answered Sep 26 '19 at 09:44

Jyrki Lahtonen

133,153

Check out this thread for more information and sources. – Jyrki Lahtonen Sep 26 '19 at 09:47
1

Bailing out with a table look-up. I am not knowledgeable enough to tell whether an elementary argument giving all the integer points of this curve exists. – Jyrki Lahtonen Sep 26 '19 at 10:04

almagest · Answer 2 · 2019-09-26T10:31:11.927

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).

Integer solution of $x^3 - x + 9 = 5 y^2$

2 Answers2