Diophantine equation $7b^2+7b+7=a^4$.

Solve in positive integers $(a,b)$ the equation $$7b^2+7b+7=a^4.$$

As the left side is divisible by $7$ we must have $7|a^4\implies 7|a$. Let $a=7c$, then the given equation reduces to $$b^2+b+1=343c^4.$$ I'm not sure how to proceed. Any hints or solutions are welcome.

Solution 1:

As I found while researching this Mathoverflow question, there is a Magma function IntegralQuarticPoints for finding all integer points on a curve $y^2 = Q(x) = ax^4+bx^3+cx^2+dx+e$ given also a rational point. Here we have $x=a$ and $Q(x) = 28x^4-147$ (the discriminant of $7(b^2+b+1) = x^4$ as a quadratic in $b$), and we know the rational point $(x,y) = (7,259)$; so we ask Magma

time IntegralQuarticPoints([28,0,0,0,-147],[7,259]);

and get output ending with

[
    [ -7, -259 ],
    [ 7, -259 ],
]
Time: 0.330

(the Time is reported in seconds). So there's a proof that the known solutions with $a=\pm 7$ are the only ones, but it might use some rather sophisticated mathematics along the way. Some of the preceding output indicates that the same equation $y^2 = 28x^4-147$ with rational (rather than integral) variables is an elliptic curve with a rational $2$-torsion point and rank $1$, which I think means that there's probably a proof that's not as advanced as whatever Magma is using but still not simple. (When a Diophantine equation has no solution at all, there is often an elementary proof; but once there's a nontrivial solution it's usually hard to find an elementary proof that allows that solution (as it must) but excludes all others.)

Solution 2:

If $$ b^2+b+1=343c^4 $$ then $b^2+b+1-343c^4=0$ $$ b=\frac{-1-\sqrt{1-4(1-343c^4)}}{2}=\frac{-1-\sqrt{1372c^4-3}}{2} $$ So there is an integral solution iff there exist $d$ such as $1372c^4-3=d^2$, that is $$ 1372c^4-d^2=3 $$ A solution for the last equation is $c=1$, $d=37$. Hence $b=18$ or $b=-19$.