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.
Best Answer
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:
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).
which gives the only positive output as $[ 1, 37 ]$