Using properties of congruence only, show that $x^2+7x+4$ is divisible by $3$ whenever $x=1+3k$ for any integer $k$.
What I have so far is:
Assume $x^2+7x+4$ is divisible by $3$. Therefore we can say $x^2+7x+4= (x^(2)+7x+4)(mod 3)= x^2(mod3)+7x(mod3)+ 4(mod3)$
Where do I go from there? Do I then plug in $1+3k$ for $x$?
Before my retirement, I taught this material many times, defining the relation $a\equiv b\pmod m$ as meaning that $b-a$ was a multiple of $m$, and the students always frowned a little, and then said that they understood.
Then I showed them how, if $a\equiv a'\pmod m$ and $b\equiv b'\pmod m$, then we had relations \begin{align} a+b&\equiv a'+b'\pmod m\\ a-b&\equiv a'-b'\pmod m\\ ab&\equiv a'b'\pmod m\,, \end{align} and they all said that they understood. And then I asked them why, since they knew that $7\equiv1\pmod3$ and $4\equiv1\pmod3$, for any integer $x$, they understood why $x^2+7x+4\equiv x^2+x+1\pmod3$, they looked at me blankly. They hadn’t understood deeply enough.
The moral of the story is that there are different levels of understanding. Yes, you “understand” when it’s shown to you the first time, but you don’t really understand till you’ve made it part of you.
The way to look at your problem is to realize that as far as congruence modulo $3$ is concerned, you’re being asked to look at $x^2+x+1$, not at the expression that was handed to you. And to realize that as far as congruence modulo $3$ is concerned, there are only three possible inputs for $x$: $x=0,1,2$. And of course only $x=1$ makes the expression zero, again modulo $3$.