Your technique should have worked, but if you don't know which expansions to do first you can get yourself in a tangle of algebra and make silly mistakes that bring the whole thing crashing down.
The way I reasoned was, well, I have four numbers multiplied together, and I want it to be two numbers of the same size multiplied together. So I'll try multiplying the big one with the small one, and the two middle ones.
$$p(p+1)(p+2)(p+3) + 1 = (p^2 + 3p)(p^2 + 3p + 2) + 1$$
Now those terms are nearly the same. How can we force them together? I'm going to use the basic but sometimes-overlooked fact that $xy = (x+1)y - y$, and likewise $x(y + 1) = xy + x$.
$$\begin{align*}
(p^2 + 3p)(p^2 + 3p + 2) + 1 &= (p^2 + 3p + 1)(p^2 + 3p + 2) - (p^2 + 3p + 2) + 1 \\
&= (p^2 + 3p + 1)(p^2 + 3p + 1) + (p^2 + 3p + 1) - (p^2 + 3p + 2) + 1 \\
&= (p^2 + 3p + 1)^2
\end{align*}$$
Tada.
In general, the contrapositive of a conditional:
'If P then Q'
is the statement:
'If not Q then not P'
Applied to your statement, we would thus get:
'If $n$ is a perfect square, then $n$ is not a positive integer such that $n \equiv 2 \pmod{4}$ or $n \equiv 3 \pmod{4}$'
... But somehow I doubt that's what they meant. In fact, the original statement was probably meant as:
'For any positive integer $n$, it holds that if $n \equiv 2 \pmod{4}$ or $n \equiv 3 \pmod{4}$, then $n$ is not a perfect square'
So then by taking the contrapositive of the contrapositive of the conditional that is part of that general statement about positive integers, we get:
'For any positive integer $n$, it holds that if $n$ is a perfect square, then it is not the case that $n \equiv 2 \pmod{4}$ or $n \equiv 3 \pmod{4}$'
...which makes a lot more sense.
Indeed, to prove this statement:
Take $n$ to be a positive integer and assume it is a perfect square. So, $n=k^2$ with $k$ an integer. $k$ is either even or odd. If $k$ is even, then $k=2m$ for some integer $m$, and so $n=(2m)^2=4m^2$. Hence, $n \equiv 0 \pmod{4}$. If $k$ is odd, then $k=2k+1$ for some integer $m$' and so $n=(2m+1)^2=4m^2+4m+1$, and hence $n \equiv 1 \pmod{4}$. So, it is not the case that $n \equiv 2 \pmod{4}$ or that $n \equiv 3 \pmod{4}$.
Best Answer
Any integer can be written as $10n+r$ where $n,r$ are integers with $0\leq r<10$. Then the unit digit of the square $$(10n+r)^2=100n+20rn+r^2=10(\underbrace{10n+2rn}_{\text{integer}})+r^2$$ is equal to the unit digit of $r^2$. So it suffices to consider the unit digit of $r^2$ when $r\in \{0,1,2,3,4,5,6,7,8,9\}$.
Thus we may conclude that the unit digit of any perfect square belongs to the set $$\{\text{unit digit of $r^2$}: r\in \{0,1,2,3,4,5,6,7,8,9\}\}=\{0,1,4,9,6,5\}.$$
P.S. As regards the original question, a product of two consecutive integers has a unit digit $6$ iff we multiply factors with unit digits $2$ and $3$, or $7$ and $8$. It follows that none of these factors can be a perfect square.