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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.

Andrés E. Caicedo
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  • 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? –  Dec 14 '14 at 20:50
  • Certainly. http://en.wikipedia.org/wiki/Quadratic_equation#Quadratic_formula Im under impression you already knew this , but perhaps ABC ( the name of the formula ) did not ring a bell. – mick Dec 14 '14 at 20:52
  • @mick Also, for your substituted equation, there are infinite solutions but I suspect that the solutions of the given equation are only $(0,1) : \text{&} : (0,-1)$. –  Dec 14 '14 at 20:53
  • @mick Oh Yes, I have used that, and that's how I factorized the given equation into the Pell-Fermat equation. –  Dec 14 '14 at 20:55
  • Let us continue this discussion in chat. –  Dec 14 '14 at 21:02
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    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
  • If I did not a mistake then $y=3z+1$ implies $(4z+1)^2-(2z)^2=(2x^2+1)^2$, so there exist integers $m,n$ such that $4z+1=m^2+n^2$, $2x^2+1=m^2-n^2$ and $2z=2mn$. – Alex Ravsky Dec 14 '14 at 22:02
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    No solutions below $10^7$. – Lucian Dec 14 '14 at 23:04
  • It seems that we can slightly simplify the problem as follows. The initial equation is equivalent to $(2y+1)(2y-1)=3(2x^2+1)^2$. Since $GCD(2y+1,2y-1)|2$, we have $2y+1=3t^2$, $2y-1=z^2$, and $tz=2x^2+1$ or $2y+1=t^2$, $2y-1=3z^2$, and $tz=2x^2+1$ for some integers $t$ and $z$. The second case is impossible, because it yields an equality $t^2\equiv 2(\operatorname{mod} 3)$. – Alex Ravsky Dec 15 '14 at 08:57
  • It my calculations are right, then I did the following. Guided by this answer I checked a recurrent sequence ${(t_n, z_n)}$ of solutions of an equation $3t^2-z^2=2$ (such that $z_{n+1}=2z_n+3t_n$ and $t_{n+1}=z_n+2t_n$) to satisfy the equality $t_nz_n=2x^2+1$ for some integer $x$. I checked it exactly from $n=0$ (where I have $t_0=1$, $z_0=1$) up to $n=16$ (where I obtained $t_{16}=1117014753$ and $z_{16}=1934726305$) and approximately up to $n=32$ (where I obtained $t_{32}=1582048049556775361$ and $z_{32}=2740187601847579969$). – Alex Ravsky Dec 15 '14 at 12:29
  • I found no solutions for the cheked non-zero values of $n$. – Alex Ravsky Dec 15 '14 at 12:33
  • The empirical evidence suggests that $7|x$ and in this case $n=0,13 (\operatorname{mod} 28)$. – Alex Ravsky Dec 15 '14 at 15:12
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    @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 Answers2

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

Community
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Kieren MacMillan
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  • 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.... –  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 02 '15 at 00:35
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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 $10^7$.

Samuel Hambleton
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    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
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    There are no integer solutions with $0< x < 1.6 \times 10^{5719}$ but my computer is slow. – Samuel Hambleton Dec 17 '14 at 05:49