Find smallest number m such that $9^{32} + 19^{433} + m$ is divisible by $4$ We went over this in class awhile ago, but I can't seem to figure out how to solve it. Obviously you can do it exhaustively with a supercomputer, but that doesn't seem practical when I know there's a simplistic way to solve it.
 A: $9\equiv 1\pmod4$, so every power of $9$ is also congruent to $1\bmod 4$; in particular, $9^{32}\equiv 1\pmod 4$. $19\equiv -1\pmod 4$, so $19^{433}\equiv(-1)^{433}\equiv -1\pmod 4$, since $(-1)^{433}=-1$. Thus, $$9^{32}+19^{433}\equiv 1+(-1)\equiv 0\pmod 4\;,$$ meaning that $9^{32}+19^{433}$ is divisible by $4$. The smallest non-negative $m$ such that $9^{32}+19^{433}+m$ is divisible by $4$ is therefore $0$, and the smallest positive $m$ is $4$.
A: Hint $\rm\,\ mod\ n\!:\ (an\!+\!1)^j + (bn\!-\!1)^{2k+1}\equiv 1^j + (-1)^{2k+1}\equiv 1-1\equiv 0$
A: I presume “the smallest number m” here is intended in the sense of “the smallest non-negative integer m”.  Assuming this, I’d suggest:


*

*first, show that for any number n, “what’s the smallest number m such that n + m is divisible by 4” is equivalent to a slightly different question about n, in a more standard form.

*secondly, use techniques about powers (which you’ve hopefully seen) to answer that question.
A: It seems you can do this simply by modding out:
$$
9^{32}\equiv 81^{16}\equiv 1^{16}\equiv 1 \mod 4
$$
$$
19^{433}\equiv (-1)^{433}\equiv -1 \mod 4
$$
Adding these, we see $9^{32}+19^{433}$ is divisible by four, so $m=0$ is valid. I assume you mean the smallest absolute value, as you can just make negative multiples of four.  Hope that's what you were looking for!
