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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 $\gcd(x,y,z)=1$?

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up vote 11 down vote accepted

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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Below is a very simple solution that avoids the (omitted) hairy arithmetic in the accepted solution. As there, reduce to $\rm\:x,y,z \not\equiv 0\:.$ Divide by $\rm\: x^3\:$ to get $\rm\: f_{a,b} = 1 + 2\ a^3 + 4\ b^3 + a\:b \equiv 0\ \ (mod\ 7)\:,$ $\rm\: a = y/x\:,\ b = z/x\:.\ $ $\rm\: n\not\equiv 0\ \Rightarrow\ n^3 \equiv \pm1\ (mod\ 7)\:$ yielding $4$ possibilities $\rm\ a^3 \equiv \pm1,\ b^3 \equiv \pm 1\:.\:$ E.g. $\:$ if $\rm\ a^3 \equiv 1,\ b^3 \equiv -1\ $ then $\rm\:f_{a,b}\:$ becomes $\rm\ a\:b \equiv 1\ $ contra $\rm\ (ab)^3 \equiv a^3\ b^3 \equiv (1)(-1) \equiv -1\:.\ $ Reasoning very similarly shows simply and quickly that the remaining $3$ cases are unsolvable.

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