Proving non-existence of rational points in a simple equation Working on Chapter 6.20 of Hammack Book of Proof

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 $k \in Z$
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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 $mod3$


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.
