1

I started by saying that $x^2$ and $y^2$ must be divisible by 9 because they are perfect squares and must add to a multiple of three. $z^2$ also must be a multiple of 9, otherwise the equation wouldn't be correct. $3z^2$ will have an extra multiple of $3$, so the equation is not true for values other than $x=y=z=0$.
Is my reasoning correct? And would this logic apply for $x^2+y^2=5z^2$?

Gerard L.
  • 2,536
  • What is your exact reasoning in your first sentence ? I don't understand why would both $x^2$ and $y^2$ be divisible by $9$, I can only deduce than $x^2 + y^2$ is divisible by $3$. Otherwise, suppose $x,y,z$ are relatively prime ; reasoning modulo $4$ will prove $x,y,z$ are divisible by 4, a contradiction. – Junkyards Sep 24 '17 at 21:33
  • Not quite, as $x^2 + y^2$ might be divisible by $27$ too. – Michael Biro Sep 24 '17 at 21:36
  • 1
    And $1^2 + 2^2 = 5 \times 1^2$, so your reasoning really can't be applied in the case $n = 5$. – Junkyards Sep 24 '17 at 21:37
  • @Michael Biro Then you can get rid of the coefficient of 3 on the RHS and simplify it down into $x^2$ and $y^2$ being divisible by 9. Am I not right? – Gerard L. Sep 24 '17 at 21:39
  • For which question? Because question 2 was solved by @Junkyards – Gerard L. Sep 24 '17 at 21:40

1 Answers1

5

Suppose $x^2+y^2=3z^2$ is a minimal non zero solution. Now consider $x$ and $y$ modulo $3$ ($1^2 \equiv 2^2 \equiv 1 \pmod{3}) $. Both must be multiples of $3$ so $x=3a$ and $y=3b$. this give $3a^2+3b^2=z^2$ so $z$ must be a multiple of $3$. $z=3c$ and we have $a^2+b^2=3c^2$ which contradict the hypothesis that $x^2+y^2=3z^2$ is a minimal non zero solution.

Donald Splutterwit
  • 36,613
  • 2
  • 26
  • 73