Proving $\gcd(n^2(n^2+1),2n+1)=\gcd(2n+1,5)$ We suppose $\forall n \in \mathbb {N}\setminus{0}$.
How can I prove that $\gcd(n^2(n^2+1),2n+1)=\gcd(2n+1,5)$?
 A: $\rm mod\,\ {2n\!+\!1}\!:\ \color{#C00}{2n\equiv-1}\,\Rightarrow\,16(n^4\!+\!n^2)\equiv (\color{#C00}{2n})^4\!\!+4(\color{#C00}{2n})^2\!\equiv (\color{#C00}{-1})^4\!\!+4(\color{#C00}{-1})^2\!\equiv\, \color{#0A0}5,\ $ therefore
$\rm\qquad (\color{#C00}{2n\!+\!1},\color{#0A0}5)=(2n\!+\!1,16(n^4\!+\!n^2)) = (2n\!+\!1,n^4\!+\!n^2)\,\ $ by $\rm\ (2n\!+\!1,16)=1\ $ and Euclid, and
since $\rm\quad\,\ (\color{#C00}a,\,\color{#0A0}b) =\, (a,\,c) \ \ if\ \ b\equiv c\:\ (mod\ a)\ \ $ [modular gcd law, heart of Euclidean algorithm]
A: \begin{align}
(2n+1)^4 - 16n^2(n^2+1) & = 32n^3 + 8n^2 + 8n+1\\
4(2n+1)^3 - (32n^3 + 8n^2 + 8n+1) & = 40n^2+16n+3\\
10(2n+1)^2 - (40n^2+16n+3) & = 24n+7\\
12(2n+1) - (24n+7) &= 5
\end{align}
Hence,
\begin{align}
\gcd(n^2(n^2+1),2n+1) & = \gcd(32n^3 + 8n^2 + 8n+1,2n+1)\\
& = \gcd(40n^2+16n+3,2n+1)\\
& = \gcd(24n+7,2n+1)\\
& = \gcd(2n+1,5)
\end{align}
A: Hints:
(1) For any $\,n\in\Bbb N\;,\;\;(2n+1,5)=1\,\,\vee\; 5\,$ ;
(2) We have 
$$ 2n+1=0\pmod 5\,\implies n=2\pmod 5\implies n^2(n^2+1)=4\cdot 5= 0\pmod 5\ldots$$
A: Using the Euclidean Algorithm, we get
$$
16n^2(n^2+1)-(8n^3-4n^2+10n-5)(2n+1)=5\tag{1}
$$
Therefore, $(n^2(n^2+1),2n+1)\mid5$.
Furthermore, since $(5,16)=1$, we have $5\mid2n+1\Longleftrightarrow5\mid n^2(n^2+1)$. Thus,
$$
(n^2(n^2+1),2n+1)=(2n+1,5)\tag{2}
$$
