TITLE: $(7a+1)x^3+(7b+2)y^3+(7c+4)z^3+(7d+1)xyz=0$ does not have integer solutions
QUESTION [6 upvotes]: 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 $\gcd(x,y,z)=1$?

REPLY [7 votes]: 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.