Just to clarify: this is in the Math Challenge II-A textbook.  The question 5.6 is the following:

The solution in the book uses both mod 4 and mod 3.  Let me rephrase the solution here. 

First of all, $k$ is an even number because $2$ is the first factor of $k$. Secondly, there is no more factor of $2$, so $k$ is not a multiple of $4$, hence $k\equiv 2\pmod{4}$.  Since a perfect square can only be $0$ or $1$ mod $4$, this means $k$ is not a perfect square itself; in addition, $k+1$ is $3\pmod 4$, also not a perfect square.

Now let's look at $k-1$: We know that $k-1\equiv 1\pmod{4}$, where we cannot conclusively say it's square or not.  Here we use mod $3$: since $k$ is at least $2\cdot 3$, it is always a multiple of $3$, so $k-1\equiv 2\pmod{3}$.  Since a perfect square is $0$ or $1$ mod $3$, $k-1$ is not a perfect square.