My Math Forum help -modular arithmetic Zn

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 April 12th, 2014, 10:14 PM #1 Newbie   Joined: Apr 2014 From: australia Posts: 17 Thanks: 1 help -modular arithmetic Zn Hi all, I've tried drawing up some addition and multiplication tables for small values of Zn, but I'm not sure if that's going to help for the question below. Prove that the only integer solution to the equation x^2 + y^2 = 3z^2 is x = y = z = 0. [Hint: first interpret this equation in Zn for an appropriate n.] any help/comments will be appreciated, thankyou!
 April 13th, 2014, 04:12 AM #2 Senior Member     Joined: Apr 2014 From: Greater London, England, UK Posts: 320 Thanks: 155 Math Focus: Abstract algebra We can prove this by contradiction. Suppose that there is a solution in which $z\ne0$. Let $z=z_0$ be the smallest positive integer in such a solution. So we have $x^2+y^2=3z_0^2$. Recall that the square of an integer not divisible by $3$ is $\equiv1\pmod3$. (An integer not divisible by $3$ is of the form $3k\pm1$ and $(3k\pm1)^2\equiv1\pmod3$.) Thus if one or both of $x$ and $y$ is not divisible by $3$, $x^2+y^2$ will not be divisible by $3$. So both $x$ and $y$ must be divisible by 3, say $x=3x_0$, $y=3y_0$. $\therefore\ 9x_0^2+9y_0^2\,=\,3z_0^2$ $\implies\ 3(x_0^2+y_0^2)\,=\,z_0^2$ So $z_0$ is divisible by 3, say $z_0=3z_1$. $\therefore\ 3(x_0^2+y_0^2)\,=\,9z_1^2$ $\implies\ x_0^2+y_0^2\,=\,3z_1^2$ This contradicts the choice of $z_0$ as the smallest positive integer $z$ in the solution of the equation. Hence we must have $z=0$, from which $x=y=0$ follows. Thanks from kp100591
 April 13th, 2014, 05:34 AM #3 Newbie   Joined: Apr 2014 From: australia Posts: 17 Thanks: 1 Thank you very very much! Thanks from Olinguito
 April 13th, 2014, 12:40 PM #4 Senior Member     Joined: Apr 2014 From: Greater London, England, UK Posts: 320 Thanks: 155 Math Focus: Abstract algebra You’re welcome. A few things of interest to mention. The proof relies on the fact that $3$ is prime and that if the sum of two integer squares is divisible by $3$ then both the integers themselves must be divisible by $3$. The same thing is true for $7$: if the sum of two integer squares is divisible by $7$ then both the integers themselves must be divisible by $7$. Hence the Diophantine equation $$x^2+y^2\,=\,7z^2$$ also has only the solution $x=y=z=0$. However the equation $$x^2+y^2\,=\,5z^2$$ has a nonzero solution: $1^2+2^2\,=\,5(1^2)$. It is possible for the sum of the squares of two integers not divisible by $5$ to be divisible by $5$. In general, given an odd prime $p$, the equation $$x^2+y^2\,=\,pz^2$$ has no nonzero solution if and only if $p\equiv-1\pmod4$. Thanks from kp100591 Last edited by Olinguito; April 13th, 2014 at 12:43 PM.

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