1
$\begingroup$

I am looking for a proof for the fact that $3n^4+3n^2+1$ can never be a perfect square for a natural number $n>0$.

I know for a fact that the statement must be true as it came up as one of the cases in a solution of the diophantine equation $y^2=x^3-1$ using the LTE lemma and, according to two different solutions I have come across, this equation has no solutions apart form $(x,y)=(1, 0)$.

I have spent a considerable amount of time looking for a suitable attack strategy, however, I was not able to make any progress. Can anyone help?

$\endgroup$
6
  • 2
    $\begingroup$ What attack strategies did you try? $\endgroup$ – Arthur Oct 29 '20 at 20:02
  • 3
    $\begingroup$ As $n=0$ works, no approach that uses only modular arithmetic can work. Hence, it must have something to do with bounding. To that end, I tried to factor $m^2=3n^4+3n^2+1$ and get some useful divisibilities (which always also lead to inequalities, of course) or to set $m=n^2+r$ with $0<r<n^2$ and then see what happens to $r$. These approaches did not work out for me, though (which does of course not mean that they are doomed altogether). $\endgroup$ – mxian Oct 29 '20 at 20:07
  • $\begingroup$ Here is a similar question $\endgroup$ – J. W. Tanner Oct 29 '20 at 20:22
  • $\begingroup$ @Sil: how did you change $k^2$ to $k^3$? $\endgroup$ – J. W. Tanner Oct 29 '20 at 20:38
  • 1
    $\begingroup$ @J.W.Tanner Using approach0 $\endgroup$ – Sil Oct 29 '20 at 20:46
1
$\begingroup$

$$3n^4+3n^2+1-k^2 =0\implies \Delta = 9-12(1-k^2)$$

Since $\Delta$ is divisible by $3$ and it is a perfect square we have $$3d^2 = 4k^2-1$$

Now $2k-1$ and $2k+1$ are relatively prime so we have $$2k-1=3a^2\;\;\;\wedge \;\;\;\; 2k+1=b^2 $$ or $$2k-1=a^2\;\;\;\wedge \;\;\;\; 2k+1=3b^2 $$

In first case we get $2 =b^2-3a^2$ and in second $2=3b^2-a^2$ (which has no solution, see by modulo 3). So we are left with $2 =b^2-3a^2$ which has something to do with the Pell equation...

$\endgroup$
1
$\begingroup$

Suppose that $3n^4+3n^2+1=m^2$, so that $$3n^4+3n^2+(1-m^2)=0$$ This is a quadratic in $n^2$. $$n^2=\frac{-3\pm\sqrt{9-12(1-m^2)}}{6}=-\frac12\pm\frac{\sqrt{12m^2-3}}6$$

So if $n$ is an integer, there is an integer $k$ such that $$12m^2-3=k^2$$ Clearly, $3|k$, so let $k=3t$ and let $s=2m$. Then we have $$3s^2-9t^2=3\\s^2-3t^3=1$$ a Pell's equation.

This is as far as I've carried it, but presumably, solving for $s$ and $t$ and putting it back in terms of $m$ and $k$ will show that you can't have an integer solution to the original equation.

$\endgroup$
0
$\begingroup$

$3n^4 + 3n^2 + 1 - x^2=0$ is an elliptic curve with Weierstrass form $s^3 + t^2 - 15s + 22=0$. According to Sage, the rank is $0$ and the torsion group has order $2$, consisting of a point at infinity (which corresponds to $n=0, x=-1$) and $s=2,t=0$ (which corresponds to $n=0,x=1$). Thus $n=0$, $x = \pm 1$ are the only rational points.

$\endgroup$
2
  • $\begingroup$ Fair enough, I am sure that this is a valid solution, but I am lacking the theoretical background to understand it and was also hoping that there would be a somewhat elementary solution. Thank you nevertheless! $\endgroup$ – mxian Oct 29 '20 at 20:22
  • $\begingroup$ @mxian: See S.Dolan’s wonderful [elementary] solution here: math.stackexchange.com/a/4004298/93271 $\endgroup$ – Kieren MacMillan Jan 30 at 17:35

Not the answer you're looking for? Browse other questions tagged or ask your own question.