Find positive integer solutions of cubic equation with three variables Find positive integer solutions of equation
$$t^3 -at^2 + bt - c=0,$$
where
$$a^3-6ab+7c=0$$
($a, b, c$ are positive integers too).
I've tried to use find solutions in modular arithmetic, but there is no simplification on this way. Then I've tried find numeric solution, but there are no solution up to 400.
Any ideas? Is there a more or less regular method for such equations?
 A: Add seven times the first equation to the second, which eliminates $c$ and gives $$7t^3-7at^2+7bt+a^3-6ab=0$$Now put $t=qa$ so that $$7q^3a^3-7q^2a^3+7abq+a^3-6ab=0$$You then either have $a=0$ (whence $b=c=t=0$) or $$(7q^3-7q+1)a^2+(7q-6)b=0$$
Try $q=1$ to see how it goes ($q$ is rational here). You get $b=-a^2$ and (from the second given equation) $c=-a^3$ and $$t^3-at^2-a^2t+a^3=0=(t-a)^2(t+a)$$
This has the solution $t=\pm a$, and you have a family of integer solutions. You can try other values of $q$.

Note: I missed the stipulation "positive" in the question - it isn't in the title. To get $b$ positive you need $7q^3-7q+1$ to have the opposite sign to $7q-6$, so you can try, for example $q=-1$ with $a^2=13b$. In the second given equation this gives $$13ab-6ab+7c=0$$ and $c=-ab$, so no solution all positive.
I haven't time just now to do more on this, but there is only a small window of values of $q$ for which the opposite sign criterion works, and it may be possible to prove that $c$ will be negative.
