Reaminder Problem 3

Note first that $(n^{2} + n + 41) \equiv (n^{2} + n + 5) \pmod{12}$ .

Next, for any integer $n$ , we have that $n = 12m + r$ for some integer m and where $0 \le r \le 11$ . As such, we see that

$(n^{2} + n + 5) \equiv (r^{2} + r + 5) \pmod{12}$ .

So it suffices to just look at the remainders for $n$ such that $0 \le n \le 11$ . (Don't worry, we'll get to the squared part later.) Doing this, we find that all remainders are one of $1, 5, 7$ or $11$ .

But for each of these values, its square is $1\pmod{12}$ . Thus for any integer $n$ , the remainder when $(n^{2} + n + 41)^{2}$ is divided by $12$ will be $\boxed{1}$ .

No formal solution in my mind as n is variable, the best way is put 1 in place of n and trust the problem.