Show That $9$ Doesn't Divide $a^2-3$

[tmt1]

For any integer $a$, show that $$9 \nmid (a^2 -3) $$

I start with

$a^2 - 3 = 9r + k$, where $k \ne 0$ and $r$ is some integer but I'm unsure how to proceed

 

[johng1]

Hi,
I must confess it took me longer to understand your starting point than to solve the problem.

[FONT=MathJax_Math]a[FONT=MathJax_Main]2[FONT=MathJax_Main]−[FONT=MathJax_Main]3[FONT=MathJax_Main]=[FONT=MathJax_Main]9[FONT=MathJax_Math]r[FONT=MathJax_Main]+[FONT=MathJax_Math]k, where [FONT=MathJax_Math]k[FONT=MathJax_Main]≠[FONT=MathJax_Main]0 and [FONT=MathJax_Math]r is some integer

I assume what you meant was:
For any integer $a$, if $a^2-3=9r+k$ for some integers $r$ and $k$, then 9 does not divide $r$. Basically, then you're talking about congruence mod 9.

For any integer $x$, 9 divides $x$ iff $x\equiv 0\pmod{9}$

So now compute all possible values mod 9 of $a^2-3$. You should quickly get the set $$\{0-3\pmod{9},\,1-3\pmod{9},\,4-3\pmod{9},\,7-3\pmod{9}\}$$

Quickly finish from here.