Find all positive integers $n$ for which $1372\,n^4 - 3$ is an odd perfect square.

I tried $\bmod ,4,5,7$ and failed. Next, I used Vieta’s Theorem and failed again.

Any hints, please. Thank you very much!

Edit number and parity already. Sorry for typo

Edit 2 : This question is related to this question.

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    $\begingroup$ I think it has to do with the pell type equation $k^2-7x^2=-3$ $\endgroup$ Jul 31, 2020 at 11:05
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    $\begingroup$ try to write $n^4=(1+(n-1))^4$ and expand it? Expand odd perfect square as (2k+1)^2, too. That may help? $\endgroup$ Jul 31, 2020 at 11:06
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    $\begingroup$ @IvartheBoneless This question appears to come from your previous one of Find all $n$ which $7(n^2 + n + 1)$ is perfect $4^{th}$ power. where multiplying both sides of your equation $n^2 + n + 1 = 343k^4$ gives $4n^2 + 4n + 4 = 1372k^4$, which becomes $(2n+1)^2 + 3 = 1372k^4$, so $1372k^4 - 3$ is equal to an odd perfect square, i.e., $(2n+1)^2$. Please provide a link & such details in the future to help people better understand the context of what you're asking, as well as to help avoid things like duplication of efforts across questions. $\endgroup$ Jul 31, 2020 at 18:34
  • $\begingroup$ Oh , sorry. @JohnOmielan $\endgroup$ Aug 1, 2020 at 6:10
  • $\begingroup$ I'm not sure any of the "answers" so far fully answers this question $\endgroup$ Aug 2, 2020 at 13:39

3 Answers 3


The equation $y^2=1372x^4-3$ has only one positive integral solution for $x$ and $y$ at which is found at $(1,37)$.

We can use the general technique in this answer https://mathoverflow.net/a/338108 to convert your quartic into Weierstrass form and then we can use MAGMA to find all integral points on the curve.

Step 1: Quartic to Cubic (Weierstrass form)

$y^2=1372x^4-3$ can be transformed into $Y^2=X^3-4116X$ using $X:=1372x^2$ and $Y:=1372xy$ via the steps below

Take $$y^2=1372x^4-3$$ Multiply both sides by $1372^2x^2$ $$1372^2x^2y^2=1372^3x^6-3\times1372^2x^2$$ $$(1372xy)^2=(1372x^2)^3-(3\times1372)(1372x^2)$$ $$Y^2=X^3-4116X$$

Step 2: Search for Integral Points

Then using MAGMA (An online version is here for you to confirm my work for yourself: http://magma.maths.usyd.edu.au/calc/) we can run the following two lines of code to find all of the integral points on our curve:

E := EllipticCurve([0,0,0,4116,0]);

And we get the result: $(0 : 0 : 1)$ which tells us that the only one solution exists (the one that we found manually $(1,37)$).

Alternatively: Easier Solution

We could also run the following to get this answer directly (I realized this command existed after doing the work above, but it confirms the same answer).

IntegralQuarticPoints([1372, 0, 0, 0, -3]);

which gives the only positive output as $[ 1, 37 ]$


The function: $\sqrt{1372n^4 - 3}$ produces one integer only: when $n = 1$, it produces $37$.

For $ n \geq 2$, the function produces decimals.

Therefore, there is no positive integer $n$ such that $1372n^4 - 3$ is a odd perfect square.

This is a brute force function that I tested out on Python - ran the code till $10000$.

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    $\begingroup$ This may be true (though $n=1$ is a positive integer so I guess you only consider $n>1$), but it's not obvious. How do you prove it? $\endgroup$
    – lulu
    Jul 31, 2020 at 10:58
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    $\begingroup$ Formatting: you need to surround the argument of the sqrt function with { } if you want to extend the radical over the whole thing. Thus...\sqrt {1372n^4-3} works. $\endgroup$
    – lulu
    Jul 31, 2020 at 11:00
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    $\begingroup$ @lulu, it's a brute force function - I tested it on Python up to $10000$ and it only produces an integer when $n = 1$. $\endgroup$ Jul 31, 2020 at 11:02
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    $\begingroup$ @lulu, thanks for the formatting help. $\endgroup$ Jul 31, 2020 at 11:02
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    $\begingroup$ Ok, but that is not a proof. Perhaps $n=10^4+1$ is a solution, who knows? I agree it is numerical evidence but it isn't a proof...you should edit your post to indicate that. $\endgroup$
    – lulu
    Jul 31, 2020 at 11:02

Just to give another way.

$1372n^4=m^2+3\Rightarrow2n^4\equiv m^2+3\pmod{10}$.

Noting $\mathbb Z/10\mathbb Z=\mathbb Z_{10}$ we have $$\mathbb Z_{10}^4=\{1,6,5,0\}\Rightarrow2\mathbb Z_{10}^4=\{2,0\}\\\mathbb Z_{10}^2=\{1,4,9,6,5,0\}\Rightarrow\mathbb Z_{10}^2+3=\{4,7,2,9,8,3\}$$

Since $2\mathbb Z_{10}^4\cap(\mathbb Z_{10}^2+3)=\{2\}$ we deduce that modulo $10$ we must have $n=1$ which corresponds to $m=7$ (because of $7^2+3\equiv2\pmod{10}$).

It follows that the only solution is $n=1$.

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    $\begingroup$ also $3^2+3\equiv2\bmod10$; and just because $n\equiv1\bmod10$ doesn't mean $n=1$ $\endgroup$ Jul 31, 2020 at 15:13
  • $\begingroup$ $n=10N+1$ but this works just for $N=0$ and corresponds with $m=10M+7$ for $M=3$ if I remember well. $\endgroup$
    – Piquito
    Aug 2, 2020 at 2:28

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