Finding the number of divisors of a number? How can I find the number of divisors of $2011\times2012\times2013\times2014+1$?
 A: $$2011\times2012\times2013\times2014+1 \\ =2012\times2013\times(2012-1)(2013+1)+1 \\
=2012\times2013\times(2012\times2013-2)+1 \\ =(2012\times2013)^2-2\times2012\times2013+1 \\ =(2012\times2013-1)^2$$
A: Hint $\ \ \ \overbrace{n(n+3)}^{\Large\ \ x_{\phantom{I_i}}}\ \overbrace{(n+1)(n+2)}^{\Large (x+2)} + 1\, =\, x(x+2)+1\, =\, (x+1)^2 =\, (n^2+3n+1)^2$
A: For any $n \in \mathbb{N}$, we have 
\begin{align}
\left(n^2+3n+1\right)^2 &= n^4+6n^3+11n^2+6n+1\\\\
&=n(n+1)(n+2)(n+3)+1
\end{align}
In particular,
$$2011\cdot2012\cdot 2013\cdot 2014+1=(2011^2+3\cdot 2011 + 1)^2 = 4050155^2$$
We can factor $4050155$ by first dividing by $5$, and then realizing with difficulty that $191$ is a factor, to get:
$$4050155=5 \cdot 191 \cdot 4241$$

This means that the number you're interested in has the form $$p^2q^2r^2$$ for primes $p,q,r$. This means the number of divisors is $$3^3=\boxed{27}$$

The number of divisors function is usually denoted $\sigma_0$, and you can read about how to calculate it given a prime factorization here. $$\sigma_0\left(p_1^{a_1}\cdot p_2^{a_2} \cdots\right) = (a_1+1)(a_2+1)\cdots$$
