<s>[INST] $(7a+1)x^3+(7b+2)y^3+(7c+4)z^3+(7d+1)xyz=0$ does not have integer solutions Let $a,b,c,d$ be integers. How can I prove that the equation
$$(7a+1)x^3+(7b+2)y^3+(7c+4)z^3+(7d+1)xyz=0$$
Does not have an integer solution $(x,y,z)$ such that $\gcd(x,y,z)=1$?
 [/INST] modulo $7$, the equation is $x^3 + 2y^3 + 4z^3 + xyz = 0$.
Cubes modulo $7$ are $0,1,-1$.
If one of $x,y,z$ is $0$, for example if $z = 0$, then you get $x^3 + 2y^3 = 0$, which is only possible if $x=y=0$.
So either $x=y=z=0$ or they are all nonzero.
If they are non zero, then $x^6 = y^6 = z^6 = 1$.
If you take $xyz = -(x^3 + 2y^3 + 4z^3)$ and cube it, you get $(xyz)^3 = -5(x^3 + y^3 + z^3)+(xyz)^3$, which implies that $x^3+y^3+z^3 = 0$. But since cubes are $1$ or $-1$, this is again impossible.
Thus the only solution modulo $7$ is $x=y=z=0$, so for any solution, $x,y,z$ have to be multiples of $7$. In particular there can't be any primitive solution, so there is no solution at all.
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