Find G. C. F. of $(8n^3 + 8n, 2n+1)$ I'm stuck with this problem, I divided $8n^3 + 8n$ by $2n+1$ and obtained $5$, so now my G. C. F is $\gcd(2n+1, -5)$.
What's next? I can't divide $2n+1$ by $-5$.
 A: Judging from the notation, it's quite likely that the exercise is not asking about polynomials, but about integers.
It's perfectly good to perform the polynomial division, because it holds for any value of $n$.
Since you found that the remainder is $-5$, you can conclude that the greatest common divisor of the two integers $8n^3+8n$ and $2n+1$ is a divisor of $5$, hence either $1$ or $5$.
It is $5$ if and only if $2n+1$ is divisible by $5$, that is, $2n\equiv4\pmod{5}$, which simplifies to $n\equiv2\pmod{5}$. Thus you have
$$
\gcd(8n^3+8n,2n+1)=
\begin{cases} 5 & n\equiv2\pmod{5} \\[4px] 1 & n\not\equiv2\pmod{5}\end{cases}
$$
A: Well, you figured out that $\gcd(8n^3 +8n, 2n+1) = \gcd(2n+1, -5)$.
And as $\gcd(\pm a, \pm b) = \gcd(a b)$ we know $\gcd(8n^3 + 8n, 2n+1) = \gcd(2n+1,5)$.
As $5$ is prime then $\gcd(2n+1, 5)$ is either $1$ or $5$.
It is $5$ if $5|2n+1$.  ANd it is $1$ if $5\not\mid 2n+1$.
And $5|2n+1 \iff$
$2n+1 \equiv 0 \pmod 5 \iff$
$2n \equiv -1 \pmod 5 \iff$
$n \equiv 2 \pmod 5$.
So the answer is:
$\gcd(8n^3+8n, 2n+1) =\begin{cases} 5 &\text{if } n\equiv 2 \pmod 5\\1&\text{if } n\not\equiv 2 \pmod 5\end{cases}$
A: Since $8n^3+8n=(2n+1)[4n^2-2n+5]-5$, we have that $\gcd(8n^3+8n,2n+1)$ divides $5$ and so is either $1$ or $5$. Both cases do occur:
For $n=1$, we get $\gcd(8n^3+8n,2n+1)=1$.
For $n=2$, we get $\gcd(8n^3+8n,2n+1)=5$.
In fact, 
$\gcd(8n^3+8n,2n+1)=5$ iff $n \equiv 2 \bmod 5$; otherwise $\gcd(8n^3+8n,2n+1)=1$.
