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I'm working on a problem that for complex $a,b,c$, when will these four numbers $a+b+c, a^2+b^2+c^2, a^3+b^3+c^3, a^4+b^4+c^4$ becomes an arithmetic progression with integer values. If we let this arithmetic progression as $u,u+d,u+2d,u+3d$ it can be verify that if the common difference $d$ is nonzero, the possible values are $$(u,d)=(0,6),(3,2),(6,4),(6,30),(7,6)$$ but I don't know is there any other solution yet.

By letting $t=u-2$, this problem eventually leading me to find all $t\in\mathbb Z$ such that $$6t^4-2t^2+4t+1$$ is a $\textbf{Perfect Square}$. I've done computing $|t|\leq 1000$ and only find out that the possible integer solutions are $t=-2,-1,0,1,4,5$, so it is reasonable to think that this expression is a perfect square for only these $6$ integer values of $t$.

It takes me to solve this Diophantine equation $N^2=6t^4-2t^2+4t+1$ but I couldn't make any progress, is there any way to solve this problem?

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    $\begingroup$ The equation $y^2=6x^4-2x^2+4x+1$ defines a genus $1$ curve. Given that it has a rational point $(x,y)=(0,1)$ IIRC this means that we have a standard process of converting it to an elliptic curve. Implying that the set of integer points is finite. If/when I find the time, I will try and recall how the process goes. We likely have users who remember the details, and know how to do it. When you have the elliptic curve in its minimal Weierstrass form, it is likely to be in the database. $\endgroup$ – Jyrki Lahtonen Oct 11 '19 at 10:39
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    $\begingroup$ I'll fear that I have to touch this elliptic curve which I don't know even a little detail (I'm a college student). It is good to know this equation has finite number of solutions in integers, I will take a good look at the database, thanks! $\endgroup$ – kelvin hong 方 Oct 11 '19 at 10:42
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    $\begingroup$ You search limit is a bit small to draw the conclusion. But in fact for $|t|\le 10^8$, there are no further solutions. By Faltings theorem, only finite many integer solutions exist and probably your list is complete. $\endgroup$ – Peter Oct 11 '19 at 10:45
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    $\begingroup$ @Peter, I think Faltings is overkill for elliptic curves. $\endgroup$ – Gerry Myerson Oct 17 '19 at 6:35
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    $\begingroup$ Are $a,b,c$ Gaussian integers ? $\endgroup$ – Yves Daoust Oct 17 '19 at 7:54
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Let the equation to be solved be $$ Q: v^2=6u^4-2u^2+4u+1 $$ This is an Elliptic Curve so there is a map for almost all the points to a Weierstrass form Elliptic Curve $E$ $$ E: y^2+a_1 xy+a_3 y=x^3+a_2x^2+a_4x+a_6 $$ For lucky cases, $E$ has only finitely many rational points so we can compute the inverse to check which ones are integral in $Q$. However in this case $E$ will have rank $1 \implies$ infinite rational points so this does not work directly.

There is a way to find the finitely many integer points for Weierstrass form Elliptic Curves. Seems like there is an adaptation of those methods (Elliptic Logarithms) for the quartic curve case. Magma has an implementation of it so you can use that for solving your problem.


1. Magma solver

Magma has a quartic solver exactly for your case. You can go to the online calculator and use the command (integer inputs)

IntegralQuarticPoints([a,b,c,d,e],[u,v])

for solving $$ V^2=aU^4+bU^3+cU^2+dU+e $$ where $[u,v]$ is a known integer point. Putting in

IntegralQuarticPoints([6,0,-2,4,1],[0,1])

nets you the solutions $(u,v)$:

[ -2, -9 ],[ 1, 3 ],[ -1, 1 ],[ 0, -1 ],[ 5, -61 ],[ 4, -39 ]

So it says that you have found all of them.

The algorithm is based on this paper. This seems to be the same methods for finding integer points on Weierstrass form Elliptic Curves but adapted for the quartic curve case.


2. Birational Equivalence to an Weierstrass form Elliptic Curve

The equation $Q$ is an Elliptic Curve so there exists a Birational Transformation to a Weierstrass form Elliptic Curve. More concretely:

Every point $(u,v), u\neq 0$ on the curve $$ Q: v^2 = 6u^4-2u^2+4u+1 $$ can be map onto $$ E: y^2+4xy=x^3-6x^2-24x+144 $$ via $$ \begin{align*} x &= \frac{(4 u + 2 (1 + v))}{u^2}, & y &= \frac{(-8 u^2 + 2 (4 u - 2 u^2) + 4 (1 + v))}{u^3} \end{align*} $$ An inverse map for $(x,y),y\neq 0$ from $E$ to $Q$ is $$ \begin{align*} u &= \frac{2 (x - 6)}{y}, &v &= \frac{(72 x - 24 x^2 + 2 x^3 + 24 y - 4 x y - y^2)}{y^2} \end{align*} $$

The transformation can be found on page 105 of this handbook.


For $u=0$, the integer points are $(0,\pm 1)$ on $Q$. For every other integer point $(u,v),u\neq 0$ there exists a map onto a point $(x,y)$ on $E$, so we can attempt to take the inverse of all rational points on $E$ (excluding $y=0$) to see which ones map to integral $(u,v)$ on $Q$.

Edit 1: if $(u,v),u\neq 0$ maps to $y=0$ then solving the transform gives $v = -1 - 2 u + 3 u^2$. Then putting back to $Q$ we get $u=0,4$.

If $rank(E)=0$ then there are only finitely many points to check, but in this case $rank(E)=1$ so there are infinitely many rational points and I am stuck here.

The condition we need is $$ u=\frac{2(x-6)}{y} $$ is integral, where $(x,y),y\neq 0$ is a rational point on $$ y^2+4xy = x^3-6x^2-24x+144 $$ but this does not seem to be sufficient to solve it directly.


Edit 2: Group Structure of $E$

The Elliptic Curve $E$ has Mordell-Weil group structure $$ \mathbb Z \times \mathbb Z/2\mathbb Z $$ Where torsion is $T = (-6,12)$ and generator $G=(12,12)$. This can be obtained from here via translation $x=X-1$ followed by $y=Y+2X$.

Therefore all the points $P$ on $E$ can be described as $$ P = [k]G\oplus [\pm 1]T $$

i.e. addition of points on Elliptic Curve

Some of the "small points" on $E$ and its reverse map to $(u,v)$ are $$ \begin{align*} G = (12 , 12) &\mapsto(1,3)\\ T\oplus G=(0, -12) &\mapsto(1,-3)\\ [2]G = (3, -3) &\mapsto(-2,9)\\ T\oplus [2]G = (6, -24) &\mapsto(0,-1)\\ [3]G = (-4, 20) &\mapsto(-1,-1)\\ [4]G = (-15/4,-39/8) &\mapsto(4,-39)\\ T\oplus [5]G = (144/25, -12/125) &\mapsto (5,61) \end{align*} $$ These generated the solutions we are looking for.

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  • $\begingroup$ Thanks! These days I've done some research on convert quartic into Weierstrass form, which I also found the same $y^2+4xy=x^3-6x^2-24x+144$ as you did. I'm curious about the algorithm the 'Magma' used. Are these six points appear because they can generated by $(0,1)$ or these six points are really all of them? And also what is the general approach will you do to find out all rational points on the Weierstrass form? I'm working on the method provided in the Elliptic curve handbook by Connell but I haven't worked things out yet. $\endgroup$ – kelvin hong 方 Oct 17 '19 at 10:13
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    $\begingroup$ I did not read the details but presumably $(0,1)$ is used for the transformation to Weierstrass form like the above. You will see that it is a requirement in the handbook. The general approach for all rational point is very difficult and not known to always work but I only know some small parts. The torsion part is simple while the generators may require computing something called canonical heights. These 6 points are all of them according to Magma and the paper. It should be correct unless there is a bug in the code. $\endgroup$ – Yong Hao Ng Oct 17 '19 at 11:51
  • $\begingroup$ It is very useful, appreciate that! $\endgroup$ – kelvin hong 方 Oct 17 '19 at 13:00
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    $\begingroup$ To find all rational points you need to find the generators $g_1,g_2,\dots, g_r$, though this can be an empty set. You also need the torsion but that part is simple. One can differentiate between different generators say by comparing their canonical heights, so the difficulty lies in (1) knowing how many there are and (2) finding them. There are (very advanced) methods that work quite often for (1) but it doesn't always work. I think it's the same for (2). $\endgroup$ – Yong Hao Ng Oct 18 '19 at 5:53

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