Prove that $x^2+2y^2+3z^2=10a^2$ has no integer solutions aside from all of them being 0

Question

I got this equation while I was trying to solve a certain math Olympiad problem. I tried modulus and whatnot, but I haven't got anywhere. Is there a way to prove this?

Answer 1

Show that all solutions mod $16$ have $x,y,z,a$ all even, and use infinite descent.

Answer 2

Solution with infinite descent.

Let $(x,y,z,a)$ be a solution with the smallest possible $a$. Say $a>0$. $$x^2+2y^2+3z^2=10a^2$$

Modulo 2 we get $2\mid x^2+z^2$.

• If $x,z$ both odd then $x=2b+1$ and $z=2c+1$ so we have $$4b^2+4b+1+2y^2+12c^2+12c+3 = 10a^2$$ so $2b^2+2b+2+6c^2+6c+y^2=5a^2$ ans thus $2\mid y^2-a^2$

  • If $y,a$ both odd, then $y=2d+1$ and $a=2t+1$ so $$b^2+b+3c^2+3c+2d^2+2d=10t^2+10t+1$$ which is impossible since left side is even and right odd.
  • If $y,a$ both even, then $y=2d$ and $a=2t$ so $$b^2+b+1+3c^2+3c+2d^2=10t^2$$ which is impossible since right side is even and left odd.

• If $x,z$ both even then $x=2b$ and $z=2c$ so we have $$2b^2+y^2+6c^2 = 5a^2$$ so $2\mid y^2-a^2$ and we have two cases again. Can you finish?

Answer 3

Just for culture:

Showing what numbers are NOT represented is the easy part. These 102 ternary forms are special (among positive forms) in that each does represent all non-excluded numbers.