Is this proof of the irrationality of the square root of $3$ valid?

Question

Assume that $\sqrt{3}=\frac{a}{b}$ where $\frac{a}{b}$ is in its simplest possible form. So $3=\frac{a^2}{b^2}$, hence $3b^2=a^2$. If $b$ is even then $a^2$ and $a$ are also even. That means they have a common factor of $2$. But this is impossible since $\frac{a}{b}$ is in its simplest form. So if $\frac{a}{b}$ does exist, $a$ and $b$ must both be odd. If $a=2k+1$ and $b=2m+1$ then $3=\frac{4k^2+4k+1}{4m^2+4m+1}$. After a few steps we get to $4(3m^2+3m)−4(k^2+k)=2$. If $3m^2+3m=c$ and $k^2+k=d$ then $4c−4d=2$ and so $4(c−d)=2$. This has no whole number solutions and that means that $a$ and $b$ are not even and not odd and that cannot be. So $\frac{a}{b}$ does not exist.

Answer 1

It looks fine to me, except for a tiny detail: after saying that if $b$ is even, then $a$ must be even too, you should have added that the reverse is true: if $a$ is even, then $b$ must be even too.

And there's no need no introduce the numbers $c$ and $d$, though. You could just say that we can't have $4(3m^2+3m-k^2-k)=2$, since $4$ doesn't divide $2$.

Answer 2

Short version:

$$c^2\bmod4=c\bmod2$$

so that with $a,b$ not both even,

$$(3a^2-b^2)\bmod4\ne0.$$