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Working on Chapter 6.20 of Hammack Book of Proof

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Show that $x^{2} + y^{2} - 3 = 0$ has no rational points.

First prove: IF $3\not\vert m$ THEN $m^{2}\equiv1(mod 3)$

If $3\not\vert m$ then either of two cases are true:

CASE $m\equiv 1 (mod3)$

$m=3k+1$ where $k \in Z$

$m^{2}= 9K^{2}+6k+1 = 3(3K^{2}+2K)+1$

Thus $m^{2}\equiv1(mod3)$

CASE $m\equiv 2 (mod3)$

$m=3k+2$ where $k \in Z$

$m^{2}= 9K^{2}+12k+4 = 3(3K^{2}+4K+1)+1 $

Thus $m^{2}\equiv1(mod3)$

So in both cases $m^{2}\equiv1(mod3)$

Therefore IF $3\not\vert m$ THEN $m^{2}\equiv1(mod 3)$

Part 2: Show general case no rational points for $a^{2}+b^{2}=3c^{2}$

$a,b,c$ are relatively prime to each other (no common factors aside 1)

By inspection, $3\vert (a^{2}+b^{2})$ thus $(a^{2}+b^{2}= 3K+0$ where $k \in Z$ ` This means $(a^{2}+b^{3})$ must have no remainder when divided by three. If $3\not\vert a$ or $3\not\vert a$ then $(a^{2}+b^{2})$ would have a remainder of $(1+0, 0+1, or 1+1)$ and violate the statement. Therefore $3\vert a$ AND $3\vert b$.

Because $3\vert a$ AND $3\vert b$ we can redefine $a=3m$ and $b=3n$ where $m,n \in Z$.

Then we rewrite $(a^{2}+b^{2})=3c^{2}$ as:

$(3m)^{2}+(3n)^2=3c^{2}$

$9m^{2}+9n^{2}=3c^{2}$

$3(3m^{2}+3n^{2})=3c^{2}$

$(3m^{2}+3n^{2})=c^{2}$

$3(m^{2}+n^{2})=c^{2}$

which means: $3\vert c^{2}$ which means $3\vert c^{2}$

This gives us a contradiction -- a, b, c all are divided by three, but we stated they were relatively prime and should only have 1 as a common factor!

Thus conclude there are no rational point solution for $a^{2}+b^{2}=3c^{2}$

final section

$a^{2} + b^{2} - 3 =0$

$a^{2} + b^{2} =3$

replace the rationals $a,b$ with rationals $p,q,m,n \in Z$

$\left(\frac{p}{q}\right)+\left(\frac{m}{n}\right)=3$

$(pn)^{2}+(mq)^{2}=3(qn)^{2}$

rename $a=pn$, $b=mq$, and $c=qn$ and we get $a^{2}+b^{2}=3c^{2}$, which we know there is no rational point solution for.

QED?

Further I'm confused by Hammack's solution/hint, as he says I should be inspecting $mod4$ results, while I believe I solved this using $mod3$

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    Sum of squares mod 4 isn't to solve but to evaluate behavior. $x^2 = 0\mod 4$ if x is even and $x^2=1\mod4$ if x is odd. So a^2+b^2 is never 3 so 3c^2 is even. So a and b are even. Devide all three by highest power of 2 and and you'll a contradiction. – 2017-02-07
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    You start talking about something called $m$ but never tell us what you mean... – 2017-02-07
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    Your proof is good. I don't see that you did the case for a^2+b^2=3c^2 if a,b,c are NOT relative prime, or alternative where you showed pn,mq,qn are relatively prime. But that's an easy and obvious minor point. But just because you are solving for multiples of 3 is no reason you need to do mod 3. Doing mod 2^2 is easier as all numbers are even or odd. 2 cases not 3. – 2017-02-07
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    You can do the same prove with x^2+y^2-19=0 and note a^2 = 0,1,4,9,16,6,17,11,7,5 mod 13 so... but then you have umpteen cases to explain. Easier to just point out a^2=0,1 mod 4. And that's incompatible with a^2+b^2=19c^2. (Which btw is equiv 3c^2 mod 4) – 2017-02-07
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    Which is kind of handy as it gives us a method of showing a^2+b^2=13c^2 *does* have solutions as 13c^2 = c^2 mod 4 so c is odd, a even and b odd so 2^2 +3^2=13*1^2 works. – 2017-02-07

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Just do the hint.

If $n=2k $ then $n^2=4k^2\equiv 0\mod 4$

If $n=2k+1$ then $n^2=4k^2+4k+1\equiv 1 \mod 4$

So $3c^2 \equiv 0|3\mod 4$

And $a^2+b^2=0,1,2\mod 4$

So if $a^2 +b^2 =3c^2$ then all $a,b,c$ are even. But if we let $a=\gcd (a,b,c)a';b=\gcd (a,b,c)b';c=\gcd (a,b,c)c'$. $a',b',c'$ can't all be even (unless they are all $0$). But $a'^2+b'^2=3c'^2$ so they must be all even. A contradiction.

So $a^2+b^2=3c^2$ has no integer solutions (except $(0,0,0) $).

Let $r=n/m;s=p/q \in \mathbb Q$ and let $(r, s) $ be a solution to $r^2+s^2-3=0$. Then $(nq)^2+(pm)^2=3 (mq)^2$ . But that is impossible.

So we aren't solving with mod 4. We are using a specific property that the sum of square integer which is true for any problem involving sums of squares.

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    sorry, one clarification i ask. Is $3c^{2}\equiv0\vert3$ mod 4 true because it's like saying "we know $c^{2}mod4$ is either 0 or 1, so 3 * 0, or 3*1 is either 0 or 3? – 2017-02-08
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    Yes, that is exactly why. If k = j \mod n, then 3k = 3j \mod n. – 2017-02-08