0

How do I prove that the equation $a^2-2b^2-3c^2+6d^2=0$ has no non trivial rational solutions?

What techniques are there to solve general problems like this?

Inspector gadget
  • 45
  • 1
  • 7
  • 1
    if there are rational solutions with some of the variables nonzero, there are integer solutions. Next, if there are integer solutions with some of $a,b,c,d \neq 0,$ then there are such with $\gcd(a,b,c,d) = 1$ Anyway, you need a degree of comfort with quadratic forms over the integers. – Will Jagy Jun 16 '23 at 20:10
  • 1
    tell you what: Since $a^2 - 2 b^2 \equiv 0 \pmod 3$ – Will Jagy Jun 16 '23 at 20:13
  • @WillJagy: what if $a$ and $b$ are both divisible by $3$? – TonyK Jun 16 '23 at 20:14
  • @TonyK right. But then $a^2 - 2 b^2 \equiv 0 \pmod 9,$ and $3c^2 - 6 d^2 \equiv 0 \pmod 9,$ so that – Will Jagy Jun 16 '23 at 20:16
  • @WillJagy: that works for me :-) – TonyK Jun 16 '23 at 20:17
  • @WillJagy unfortunately, I have no degree of comfort of quadratic forms over integers. Could you point in the direction of literature? – Inspector gadget Jun 16 '23 at 20:23
  • inspector, I like Cassels, Rational Quadratic Forms. The thing you need to be able to prove is that, for integers $u,v,$ if $u^2 - 2 v^2 \equiv 0 \pmod 3,$ then both $u,v \equiv 0 \pmod 3,$ so that $3 | \gcd(u,v)$ and actually $u^2 - 2 v^2 \equiv 0 \pmod 9.$ This is general for binary quadratic form $e u^2 + f uv + g v^2$ with discriminant $\Delta = f^2 - 4eg,$ given a prime $q$ that does not divide $\Delta,$ if Legendre $ (|\Delta|q) = -1$ and $e u^2 + f uv + g v^2 \equiv 0 \pmod q,$ then both $u,v \equiv 0 \pmod q$ – Will Jagy Jun 16 '23 at 20:26
  • @WillJagy So if we've shown that $a^2-2b^2\equiv 0 (mod 3)$ and $ (mod 9)$ how does this help to prove there are no non trivial integer solutions? – Inspector gadget Jun 16 '23 at 20:52
  • then $3c^2 - 6 d^2 \equiv 0 \pmod 9,$ that is $c^2 - 2 d^2 \equiv 0 \pmod 3,$ so that $c,d \equiv 0 \pmod 3$ Thus $3|\gcd(a,b,c,d)$ This contradicts the original assumption that $\gcd(a,b,c,d) = 1$ – Will Jagy Jun 16 '23 at 20:55
  • Essentially it's Fermat's method of infinite descent. (Rephrasing Will's comments) We have $a^2 - 2b^2 = 3(c^2 - 2d^2)$ which implies $ a = 3a_1, b = 3b_1$ (which we can show by checking values mod 3) and so $ 3(a_1^2 - 2b_1^2 ) = c^2 - 2d^2$ which implies $ c = 3c_1, d = 3d_1$ and so on. – Calvin Lin Jun 16 '23 at 21:22

0 Answers0