Is $2^{218!} +1$ prime? 
Prove that $2^{218!} +1$ is not a prime number.

I can prove that the last digit of this number is $7$, and that's all.
Thank you.
 A: $$218!=3n:\;\;\;2^{218!}+1=(2^{n}+1)(4^{n}-2^{n}+1)$$
A: Another one: $2^{2^{213}}+1\mid2^{218!}+1$ because $218!=2^{213}\cdot k$ with $k$ odd.
A: Well, $a^3+1=(a+1)(a^2-a+1)$, so it's not prime, and we have
$$2^{218!}=(2^{2\cdot4\cdot5\cdot..\cdot 218})^3\,.$$
A: Hint $\ $ If $\rm\: k\:$ is odd then $\rm\:a^n\!+\!1\mid a^{nk}\!+\!1\ $ by $\rm\ mod\ a^n\!+\!1\!:\ a^n\!\equiv -1\:\Rightarrow\:a^{nk}\!\equiv (a^n)^k\!\equiv (-1)^k\equiv -1.\:$
Or $ $ Factor Theorem $\rm\:\Rightarrow\: x\!-\!c\mid x^k\!-\!c^k\: $ in $\rm\:\Bbb Z[x],\:$ so $\rm\:c=-1\:\Rightarrow\: x\!+\!1\mid x^k\!+\!1,\:$ hence evaluating at $\rm\:x = a^n\:$   yields $\rm\:a^n\!+\!1\mid a^{nk}\!+\!1\:$ in $\,\Bbb Z.\:$ Thus this integer divisibility results is a special case of a polynomial divisibility result. Factors of this form are sometimes called algebraic factors.
Similar to the example above, often number identities are more perceptively viewed as special cases of function or polynomial identities. $ $ For example, $ $ Aurifeuille, Le Lasseur and Lucas $ $ discovered so-called Aurifeuillian factorizations of cyclotomic polynomials $\rm\;\Phi_n(x)\, =\, C_n(x)^2\! - n\, x\,  D_n(x)^2\;$. These play a role in factoring integers of the form $\rm\; b^n \pm 1\:$, cf. the Cunningham Project. Below are some simple examples of such factorizations.
$$\begin{eqnarray}
\rm x^4 + 2^2 &=\,&\rm  (x^2 + 2x + 2)\;(x^2 - 2x + 2) \\\\
\rm \frac{x^6 + 3^3}{x^2 + 3} &=\,&\rm  (x^2 + 3x + 3)\;(x^2 - 3x + 3) \\\\
\rm \frac{x^{10} - 5^5}{x^2 - 5} &=\,&\rm  (x^4 + 5x^3 + 15x^2 + 25x + 25)\;(x^4 - 5x^3 + 15x^2 - 25x + 25) \\\\
\rm \frac{x^{12} + 6^6}{x^4 + 36} &=\,&\rm  (x^4 + 6x^3 + 18x^2 + 36x + 36)\;(x^4 - 6x^3 + 18x^2 - 36x + 36) \\\\
\end{eqnarray}$$
For more on this and related topics see Sam Wagstaff's  introduction to the Cunningham Project. 
A: In general, we the following to be true.
Claim:
If $a^n+1$ is a prime, then $n = 2^k$. To put it the other way around: If $n = 2^k \cdot m$, where $m$ is odd and greater than $1$, $a^n+1$ is composite.
Proof:
Let $k$ be the highest power of $2$ such that $2^k \vert n$, i.e., $2^k \Vert n$. We then have $n = 2^k \cdot m$, where $m$ is odd. Hence, we have
$$a^n+1 = a^{2^k \cdot m} + 1 = \left(a^{2^k}\right)^m + 1$$
Now we have
$$b + 1 \vert b^m+1$$whenever $m$ is odd. Hence, we have $$a^{2^k}+1 \vert \left(a^{2^k}\right)^m + 1 \implies a^{2^k}+1 \vert a^n+1$$ Hence, if $m \neq 1$, we have $a^n+1 > a^{2^k}+1$ and $a^{2^k}+1 \vert a^n+1$. Hence, if $m \neq 1$, we get that $a^n+1$ is not a prime. This forces $m$ to $1$. Hence, if $a^n+1$ is a prime, then $n = 2^k$. 

In your case, take $a=2$ and $n = 218!$ and clearly (Why?) $218! \neq 2^k$, where $k \in \mathbb{Z}^+$. Hence, $2^{218!}+1$ is not a prime.
A: $x=1 \cdot 2 \cdot 4 \cdot 5 \dots 218$
$2^{218!}+1 =2^{3x}+1=(2^{x}+1)(2^{2x}-2^{x}+1)$
A: $$218! \equiv 0 \mod \left( \left\{3,5,7,9,11,13,15,...,217!! \right\}\right)$$

$$\left(2^{\frac {218!}{2k+1}} \right)^{2k+1}+1 \equiv 0 \left(\mod 2^{\frac {218!}{2k+1}}+1 \right)$$, where $k≤\frac {217!!+1}{2} .$

