A Diophantine Equation related to arithmetic progression: $T_n=a^n+b^n+c^n$. 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?
 A: 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.
