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

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.

• I think it has to do with the pell type equation $k^2-7x^2=-3$ Jul 31, 2020 at 11:05
• try to write $n^4=(1+(n-1))^4$ and expand it? Expand odd perfect square as (2k+1)^2, too. That may help? Jul 31, 2020 at 11:06
• @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. Jul 31, 2020 at 18:34
• Oh , sorry. @JohnOmielan Aug 1, 2020 at 6:10
• I'm not sure any of the "answers" so far fully answers this question Aug 2, 2020 at 13:39

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]);
IntegralPoints(E);


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 $$10000$$.

• 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?
– lulu
Jul 31, 2020 at 10:58
• 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.
– lulu
Jul 31, 2020 at 11:00
• @lulu, it's a brute force function - I tested it on Python up to $10000$ and it only produces an integer when $n = 1$. Jul 31, 2020 at 11:02
• @lulu, thanks for the formatting help. Jul 31, 2020 at 11:02
• 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.
– 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 $$n=1$$.

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