5

I want to find the integer solutions to this Diophantine equation: $$5x^3=y^2+1$$ I have seen a lot of problems with monic variables, but not with a constant on the $x^3$ such as this.

I know I can factorise the right hand side and get $5x^3=(y-i)(y+i)$, so I can work in $\mathbb Z[i]$. But I am unsure where to proceed from here, and how the $5$ comes into the problem.

Parcly Taxel
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MathsIsFun
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  • I added a second problem to the question, I thought if there was an accepted solution it wouldn't be highlighted with a problem still to be answered. – MathsIsFun Apr 14 '20 at 14:26
  • Right, I updated my answer to solve the other question as well. – Parcly Taxel Apr 14 '20 at 15:03
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    Please do not edit your question to change its meaning or to add extra problems, particularly after you have gotten an answer to your original question. If you have another question, please use the "Ask a Question" dialog to post that question. – Xander Henderson Apr 14 '20 at 23:02
  • @XanderHenderson Thank you for your comment. I have now edited the question to provide further clarity to the way I have been taught how to solve problems such as these. I added an extra problem in order to try and get a solution using the UFD method, as mentioned in the comments below. Please can you consider reopening my question, as I still do not know how to solve this using UFD's. – MathsIsFun Apr 15 '20 at 11:34
  • @MathsIsFun You thank me for my comment, then proceed to utterly ignore it. Please do not edit your question to ask a new question once you have an answer. Please use the "Ask Question" button at the top of the page to ask a new question. – Xander Henderson Apr 15 '20 at 14:52
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    I'm sorry I don't see how I have ignored it? You closed my question, and it said to edit it to give further clarity in order to have it reopened. My question was not answered, not in a way I know how to do, as has been made clear further down. I have found that if I ask a question similar to this, it would be marked as a duplicate which seems pointless. You criticized my question for being unclear, which it was, I have now made it clear and am awaiting an answer in the method I have specified, which is how I have made my question clear. – MathsIsFun Apr 15 '20 at 15:17
  • I always wanted my question answered in the method using UFD's, hence why I factorised it that way. However, as you made obvious this wasn't clear. I have now made it clear in the question. As you can see in the comments further down, I asked if this could be solved this way, I was told no. I added more detail to try and understand how to use the method I was taught. It is now clear it can be solved the method I was asking, hence I worked to get it reopened so I can understand this method further. – MathsIsFun Apr 15 '20 at 15:19
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    (1) I did not close your post. I was one of five users who voted to close your post. (2) You asked a question and then got an answer. After getting that answer, you determined that it was not the answer you wanted (because your question was unclear). You then edited the question to ask something that appeared to be quite different. I reverted that edit, and asked you to post a new question. You then edited your question again to ask a different question. Once you get an answer, you should not edit your question to change its meaning. Ask a new question. – Xander Henderson Apr 15 '20 at 15:27
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    I have asked the question regarding Gaussian integer-based proofs here. – Parcly Taxel Apr 15 '20 at 15:37

1 Answers1

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If we multiply both sides by $25$ and rearrange we get $$(5y)^2=(5x)^3-25$$ So an integral solution to the original solution corresponds to an integral point on the elliptic curve $w^2=z^3-25$. But according to the LMFDB there are only two integral points on the curve, $(z,w)=(5,\pm10)$. We conclude that the only integral solutions to the original equation are $(x,y)=(1,\pm2)$.

Letting $z=Nx,w=Ny$ shows that integral solutions to $Nx^3=y^2+1$ for fixed $N$ correspond to integral points on the Mordell curve $w^2=z^3-N^2$, so there are always finitely many solutions and they may be rather effectively counted. In the $N=17$ case, the only solutions are $(x,y)=(1,\pm4)$.

Parcly Taxel
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  • Thank you for your quick response. I wonder, is there a way to solve it using the factorisation I made above? – MathsIsFun Apr 14 '20 at 10:48
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    @MathsIsFun No. – Parcly Taxel Apr 14 '20 at 10:54
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    Are you sure? The Gaussians are a UFD, $y$ must be even, $\gcd(y+i,y-i)=1$, so $y+i=\alpha(a+bi)^3$ where $\alpha$ is a unit or an associate of $2\pm i$ or of $5$, and so on – occasionally, that approach works. – Gerry Myerson Apr 15 '20 at 04:39
  • @GerryMyerson Where is $x$ in your equations? – Parcly Taxel Apr 15 '20 at 04:50
  • $(y+i)(y-i)=5x^3$, so $y+i$ is a cube, up to factors of $5$. – Gerry Myerson Apr 15 '20 at 06:38
  • @GerryMyerson I would definitely be interested to see your suggestion in more detail, if you're happy to provide one? I haven't solved questions like this using elliptical curves, only using UFD's etc. – MathsIsFun Apr 15 '20 at 11:31
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    @Servaes That was my first instinct. I took the elliptic curve idea from the answer to this question of mine, and once I read the answer I was hooked. I do acknowledge now that there can be answers using the Gaussian integers, but I consider the elliptic curves a more intuitive fit for the problem. – Parcly Taxel Apr 15 '20 at 12:17
  • @ParclyTaxel To me there is nothing intuitive to a solution that boils down to "According to LMFDB...", but I'm glad that you at least acknowledge that it can be done by elementary methods in $\Bbb{Z}[i]$. – Servaes Apr 15 '20 at 12:23
  • @Servaes It's good to know this can be done using elementary methods in $\mathbb{Z}[i]$. If this question is reopened would you be happy to post the solution? – MathsIsFun Apr 15 '20 at 12:44
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    Four votes to reopen. @Servaes we may get to see your solution! – Gerry Myerson Apr 15 '20 at 12:58
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    @GerryMyerson I cast the last reopen vote. – Parcly Taxel Apr 15 '20 at 12:59
  • Thank you so much @ParclyTaxel! – MathsIsFun Apr 15 '20 at 13:00
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    @Servaes ball's in your court. Go for it! – Gerry Myerson Apr 15 '20 at 13:00
  • @GerryMyerson Would you be happy to provide the beginnings of this solution, if Servaes is unable to? – MathsIsFun Apr 15 '20 at 15:24
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    @MathsIsFun I am going to have to ask the question on your behalf. – Parcly Taxel Apr 15 '20 at 15:25
  • @ParclyTaxel Thank you, I appreciate that. All I want to do is understand! – MathsIsFun Apr 15 '20 at 15:26
  • I think I did provide the beginnings of a possible solution in my earlier comments. I haven't worked it through to see whether it actually leads to a solution. – Gerry Myerson Apr 15 '20 at 22:41
  • Here is the question @ParclyTaxel has posted: https://math.stackexchange.com/questions/3626900/integral-solutions-to-nx3-y21-by-gaussian-integers – Gerry Myerson Apr 15 '20 at 22:42
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    Not convinced my idea leads to an elementary solution. I got as far as $y+i=(2+i)(a+bi)^3$ which yields $a^3+6a^2b-3ab^2-2b^2=1$. There are techniques for handling such cubic forms, but the use more than just the arithmetic of the Gaussian integers (and they aren't in my toolbox). – Gerry Myerson Apr 17 '20 at 00:36