prove that there are no non-trivial solution to x^3+ 2y^3+4z^3=9w^3.?

Question

Thanks Curt Monash. Your ideas are a great help. taking mod 9 is the key.

Assume a solution exist. we want a primitive solution (x,y,z,w)=1. This implies that 3 divides at most two of x,y,z,w. We see that any number congruent to n is congruent to a,2,3,4,5,6,7,8(mod9). so n^3 is congruent to 0,1,8(mod 9). or we could say 0,1,-1(mod 9). take our original equation:x^3+2y^3+4z^3=9w^3. X^3==0,1,-1(mod 9) =0(mod9).
2y^3==0,2,-2(mod9).
4z^3 ==0,4,-4(mod9).

this implies that x|x^3 and x|y^3 and x|z^3, which implies 3|x, 3|y, and 3|z. implies 3|w.
this then is the contradiction, because we had already say that 3 can only divide at most two of x,y,z,w. So our original assumption is correct.

Answer 1

I don't immediately see a proof, but I know how I'd go about looking for one.

I'd look at the equation modulo 2. Then maybe modulo 4 and even 8.

And certainly modulo 3, since the exponents are 3. And maybe even modulo 9.

And I'd hope that that got me most or all of the way to a proof by contradiction.

That's if this is for an elementary course. If you're studying modular forms, elliptic functions, and so on you're out of my league. I stopped at commutative algebra.