# How to solve $y^2=3x^4+3x^2+1$ for integers.

If $x,y \in \mathbb Z$ , then find all the solutions of
$$y^2=3x^4+3x^2+1$$

I was asked this question by my friend who said that he encountered this while solving another problem. I have tried several things but am unable to solve this question. Moreover, this has to be done using elementary methods only. So far, I have tried to factorize and use Pell's equation. At the end, I'm getting
$$2y_{n} + (2x^2_{n}+1)\sqrt{3}=(2+\sqrt{3})^{n}$$

where $n \in \mathbb Z^{+}$

But I'm not able to figure out how to show a contradiction from here. Can anyone please help me out?
Thanks.

• Have you considered solving $Y = 3 X^2 + 3 X + 1$ ? Because then the square solutions for Y and X are what you want. ( $y^2 = Y , x^2 = X$ ) For $Y = 3 X^2 + 3 X + 1$ you might want to consider using the ABC formula. – mick Dec 14 '14 at 20:48
• @mick Pardon me, but can you please elaborate or give a link about the ABC formula? – user196761 Dec 14 '14 at 20:50
• Note that the given equation is equivalent to $y^2+x^6=(x^2+1)^3$. I believe the solutions to $y^2+x^6=z^3$ have been parametrized; one could locate that parametrization and see if ever $z=x^2+1$. – Greg Martin Dec 14 '14 at 21:27
• No solutions below $10^7$. – Lucian Dec 14 '14 at 23:04
• @AlexRavsky It is easy to show that since $y$ must be odd, $x$ must be divisible by $4$ (work modulo $8$). – Mark Bennet Oct 29 '20 at 20:45

## 2 Answers

The only solutions are $(x,y)=(0,-1)$ and $(x,y)=(0,+1)$. I gave an elementary proof here.

• Hey! You mentioned in your answer that there is a flaw that needs to be rectified. Anyways, I still wonder whether it can be solved by Pell-Fermat method.... – user196761 Feb 27 '15 at 17:14
• I corrected that flaw in the current edit — I think it's valid as is. And, yes, it can be solved using Pell equation as well! (MathGod sent me such a proof by email.) – Kieren MacMillan Mar 2 '15 at 0:35

This does not immediately answer your question but here is how I would approach it. Let $X = 4 y$ and $Y = 2 x^2 + 1$. Then there exists integers $x, y$ satisfying $y^2 = 3 x^4 + 3 x^2 + 1$ if there exists an integer solution $(X, Y)$ to $$X^2 - 12 Y^2 = 4$$ satisfying $4 \mid X$ and $Y$ is odd of the form $2 x^2 +1$. We know the fundamental solution is $(X, Y) = (4, 1)$. The other solutions can be generated by the polynomials $f_n(X)$ and $g_n(X)$ by $(X_n , Y_n) = (f_n(X), Y g_n(X))$, where $f_{-1}(X) = X$, $f_0(X) = 2$, $g_{-1}(X) = -1$, $g_0(X) = 0$, $$f_{n+1} = X f_{n} - f_{n-1},$$ $$g_{n+1} = X g_{n} - g_{n-1},$$ and of course $X = 4$, $Y = 1$. It is easy to show that $n$ must be odd. Now define $F_1 = 1$, $F_3 = X + 1$, $$F_{2 k + 3} = X F_{2 k + 1} - F_{2 k - 1}.$$ These are incidentally a lot like cyclotomic polynomials. Check that $F_n(X^2 - 2) = g_n(X)$. Your question requires $\frac{F_n(14) -1}{2}$ to be a perfect square. Congruences of $\frac{F_n(14) -1}{2}$ modulo primes can show that the odd number $n$ must satisfy certain congruence conditions. If $y^2 = 3 x^4 + 3 x^2 + 1$ has a solution $\not= (0, 1)$, then it must be huge, much larger than $10^7$.

• Or: The sequence $z_1,z_2,\ldots$ of non-negative integers $z$ for which $3z^2+3z+1$ is a perfect square is given by the recurrence $z_1=0$, $z_2=7$, and $z_n=14z_{n-1}-z_{n-2}+6$. – Aravind Dec 16 '14 at 14:56
• There are no integer solutions with $0< x < 1.6 \times 10^{5719}$ but my computer is slow. – Samuel Hambleton Dec 17 '14 at 5:49