Exercise 5.4.7 (Equation $f(r,s,t) + 7 f(u,v,w) + 49 f(x,y,z) = 0$, where $f(x,y,z) = x^3 + 2y^3 + 4z^3 - 6xyz$)

Proof. Assume for the sake of contradiction that $f(r,s,t)+7f(u,v,w)+49f(x,y,z)=0$, where $(r,s,t,u,v,w,u,v,w,x,y,z)\ne (0,0,0,0,0,0,0,0,0)$. Then $d=r\wedge s\wedge t\wedge u\wedge v\wedge w\wedge x\wedge y\wedge z\ne 0$, so there are integers $R,S,T,U,V,W,X,Y,Z$ such that

Since $f$ is homogeneous of degree $3$, we obtain after division by $d^3$,

Reducing this equation modulo $7$, we obtain $f(R,S,T)\equiv 0(\mathit{mod}7)$. By Problem 6,

Therefore $7^3\mid f(R,S,T)$, thus $7^3\mid 7f(U,V,W)+49f(X,Y,Z)$, so $7^2\mid f(U,V,W)+7f(X,Y,Z)$. This shows that $7\mid f(U,V,W)$. By the same Problem 6,

Therefore $7^3\mid f(R,S,T)$, thus $7^3\mid 7^{2}f(X,Y,Z)$, so $7\mid f(X,Y,Z)$, which gives

Then $7\mid R\wedge S\wedge T\wedge U\wedge V\wedge W\wedge X\wedge Y\wedge Z=1$. The contradiction $7\mid 1$ shows that the equation $f(r,s,t)+7f(u,v,w)+49f(x,y,z)=0$ has no nontrivial integral solution. □