If $ n$ is an integer and $3 | (n-4)$ then $3|[n^2-1]$? I'm getting stuck on this problem. So far I got if $3|(n-4)$ then $3 = (n-4)k$ where k is an integer then [(3/k) + 4]^2 = n. If I sub that into $[(n^2) -1]$ I get $[((3/k) + 4)^2 -1]$, which gives me $(9/k^2) + 24/k +15$ and if I pull out a $3$ I get $3[(3/k^2) + 8/k + 5]$, but that doesn't show that it multiplied by an integer equal to $3$.
 A: $3 \mid (n-4) \implies 3 \mid (n-1) \implies 3\mid (n-1)(n+1) \implies 3 \mid n^2-1  $
A: 
I'm getting stuck on this problem. So far I got if 3|(n−4) then 3=(n−4)k where k is an integer

You have that backwards
If $3|n-4$ then means $n-4 = 3k$ where $k$ is an integer.

then [(3/k) + 4]^2 = n

Why did you square it?  
If we solve for $n$ then$n = 3k + 4$.

If I sub that into [(n2)−1] I get [((3/k)+4)2−1]

Okay, I think that square was a typo.
Subbin $3k+4$ into $n^2 -1$ we get $n^2-1 =(3k+4)^2 -1 = 9k^2 + 24k + 16 -1 = 9k^2 + 24k + 15$
And pull out a $3$ we get
$=3(3k^2 + 8k + 5)$.
Which works just fine.
.....
But it's easier.
If $a|b$ then $a$ will divide $b$ plus or minus any multiple of $a$
So if $3|n-4$ then $n$ will divide $n-4 + 3 = n-1$.
ANd if $a |b$ and $b|c$ then $a|c$.  (if that isn't obvious:  If $b= ka$ and $c= bj$ then $c = bj = a(kj)$ and if $k,j$ are integers so is $kj$.)
And $n^2 - 1 = (n+1)(n-1)$.  (That's a formula that should eventually be automatic everytime you see it.)
So $3|n-1$ and $n-1|n^2-1$, so $3|n^2-1$.
... well, it was easier for me.  But eventually such things should become second nature.
A: If $3|n-4$ then $3|n-4+3=n-1$. And: $n^2-1=(n-\cdots)(n+\cdots)\cdots$
A: By Polynomial remainder theorem  evalution at $n=4$ leads to a remainder on division by $n-4$ of 15, which 3 divides. It then follows by factoring out, that this causes a factor of 3 in the original polynomial.
A: If $3|n-4$, then $n\equiv 4[3]$ so that $n^2\equiv 16\equiv 1[3]$ and thus $3|n^2-1$.
