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I'm attempting to work through some of the questions in Whitelaw's "Introduction to Abstract Algebra" but am having some difficulty.

The question is as follows

Show that $\forall n \in \Bbb{Z},$ $n^2 \equiv 0\ \text{or}\ 1 \mod{3}.$ Show further that $$3\rvert(x^2 + y^2) \implies 3\rvert x\ \text{and}\ 3\rvert y.$$ Deduce that there is no solution in $\Bbb{Z}$ to the equation $$x^2 + y^2 = 3z^2$$ except $x = y = z = 0.$

For the first part, I simply let $n = 3k,$ $3k+1$ or $3k + 2$ with $k \in \Bbb{Z}$. Hence,

\begin{align} (3k)^2 &\equiv 0 \mod{3}\\ (3k+1)^2 = 9k^2 + 6k + 1 &\equiv 1\mod{3}\\ (3k+2)^2 = 9k^2 + 12k + 3 + 1 &\equiv 1\mod{3}. \end{align}

For the second part, since $n^2 \equiv 0$ or $1 \mod{3}$, and that the only case in which $n^2 \equiv 0 \mod{3}$ was when $3\rvert n$, it must then be the case that both $x^2$ and $y^2$ are divisible by $3$ for $3$ to divide their sum (else we get that $x^2 + y^2 \equiv 1$ or $2 \mod{3}$). We also know that if $3\rvert n$ then $3\rvert n^2$. Hence, $3\rvert x$ and $3\rvert y$.

I am, however, unsure how to approach the third part.

Thanks for any help!

Edward Evans
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    Comment on the proof on the second part: If $3|n$ then $3|n^2$, but you want the converse of that statement. The only way that $3|(x^2+y^2)$ is if $x^2$ and $y^2$ are both $\equiv 0 \mod 3$. Now, use (from what you computed) that $n^2\equiv 0 \mod 3$ if and only if $n\equiv 0 \mod 3$. The idea was correct, but reorganizing might make it a touch clearer. – Nicholas Stull Apr 16 '16 at 01:04
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    Your very last sentence doesn't follow. If 3|n then 3|n^2, yes, but the converse doesn't follow.(9|36 but 9 doesn't divide 6) Here we need to state that 3 is prime and if 3|n^2 then 3|n. And that should give you a hint to the second part. Everything up to that last sentence was perfect. – fleablood Apr 16 '16 at 01:07
  • I see, thanks for your help. I will amend this in my notes! – Edward Evans Apr 16 '16 at 01:08
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    x^2 + y^2 = 0 mod 3 only if both x and y are divisible by 3 as x^2 and y^2 can only be = 0,1 (not -1) mod 3. So if x^2 + y^2=3z^2 then 3 divides x and y. So 9 divides x^2 +y^2 so 3 divides z^2 so 9 divides z^2. so 27 divides x^2 and y^2 and so on ad infinitim. – fleablood Apr 16 '16 at 01:13
  • @fleablood, I forgot to add that $3$ is prime, which is a more elegant way of using $3|n^2|$ to establish that $3|n$. – Nicholas Stull Apr 16 '16 at 01:20

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Let's suppose that we have a solution $(a,b,c)$ where $c\neq 0$ is the smallest positive integer such that the equality holds.

$$a^2 + b^2= 3c^2 \Rightarrow 3|(a^2 + b^2)$$

But we know that this means $3|a$ and $3|b$. However, on squaring we get that $9$ divides the LHS, and so it must also divide the RHS: $$a'^2 + b'^2 = \frac{c^2}{3}$$

Where $3a' = a, 3b' = b$. But $3$ is prime, so if it divides the product it must divide $c$. Factor that out as $3c'$ and divide: $$a'^2 + b'^2 = 3c'^2$$

But this is another solution triple with $c' < c$ (and also $c' \neq 0$), and we chose the smallest possible $c$, so we have a contradiction.