# 1957 AHSME Problems/Problem 32

## Problem

The largest of the following integers which divides each of the numbers of the sequence $1^5 - 1,\, 2^5 - 2,\, 3^5 - 3,\, \cdots, n^5 - n, \cdots$ is:

$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 60 \qquad \textbf{(C)}\ 15 \qquad \textbf{(D)}\ 120\qquad \textbf{(E)}\ 30$

## Solution

To begin, we look for a maximum value that this common divisor could have by looking at the smallest terms. $1^5-1=0$, but this gives us no information because all positive integers divide zero. However, $2^5-2=30$, so our answer cannot be greater than $30$ (because $30$ divided by an integer larger than $30$ is never an integer). This fact eliminates options (B) and (D).

Now, we will factorize $n^5-n$ to discover some properties: \begin{align*} n^5-n &= n(n^4-1) \\ &= n(n^2-1)(n^2+1) \\ &= n(n-1)(n+1)(n^2+1) \end{align*} One of $n$ and $n-1$ must be even, so $2$ divides $n^5-n$ for all $n$. Likewise, one of $n-1$, $n$, and $n+1$ must be divisible by $3$, so $3$ must divide $n^5-n$ as well. In the case of $5$, there are, in total, three cases:

Case 1: $5$ divides either $n-1$, $n$, or $n+1$. Then, $5$ clearly must divide $n^5-n$.

Case 2: $5$ divides $n+2$. In this case, $n+2 \equiv 0$ $\mod$ $5$, so $n \equiv 3 \mod 5$. Then, $n^2+1 \equiv 10 \equiv 0 \mod 5$, so $n^2-1$ (and, consequently, $n^5-n$) is divisible by $5$.

Case 3: $5$ divides $n+3$. In this case, $n+3 \equiv 0 \mod 5$, so $n \equiv 2 \mod 5$. Thus, $n^2+1 \equiv 5 \equiv 0 \mod 5$, so $n^2-1$ (and, consequently, $n^5-n$) is divisible by $5$.

Thus, 5 divides $n^5-n$ for all $n$.

Because $2$, $3$, and $5$ all divide $n^5-n$ and are relatively prime, $2 \cdot 3 \cdot 5 = 30$ also divides $n^5-n$. Thus, because $30$ is the largest of our remaining choices, we choose answer $\boxed{\textbf{(E) }30}$.