Spivak Calculus 4-th Ed., Chapter 2, Exercise 13a, Understanding the proof of $\sqrt3$ being irrational.

Let $k$ be an integer. It can take one of the following forms: $3n$ or $3n+1$ or $3n+2$ where $n$ is simply the quotient in the euclidean division of $k$ by $3$.

Now:

• if $k=3n$ then $k^2 = 9n^2$
• if $k=3n+1$, then $k^2 = 3\times ... +1$ (we don't really care of the exact value of $...$), thus $k^2$ isn't multiple of $3$
• if $k=3n+2$, then $k^2 = 3\times ... +1$, thus $k^2$ isn't multiple of $3$.

Thus, if $k^2$ is multiple of 3, then $k$ can't be of the form $3n+1$ or $3n+2$. Hence, it must be of the form $3n$, that is, $k$ must be multiple of 3.

Hence, if $k^2$ is multiple of $3$, then $k$ is also multiple of $3$.

The state ment

If $k^2$ is divisible by $3$ then so is $k$.

holds for all integers $k$. He uses this fact below for $k=p$ and for $k=q$.

In the first part, Spivak proves that of a number is not a multiple of $3$ (that is, if it is of the type $3n+1$ or $3n+2$), then its square is not a multiple of $3$. Therefore, if the square is a multiple of $3$, then the number itself must be a multiple of $3$ too.