Putnam 2001, Problem A-5

Prove that there are unique positive integers $a, n$ such that $a^{n+1} - (a+1)^n = 2001$.

Solution:
Suppose $a^{n+1} - (a+1)^n = 2001$. Notice that $a^{n+1} + [(a+1)^n - 1]$ is a multiple of $a$; thus $a$ divides $2002 = 2 \times 7 \times 11 \times 13$.

Since 2001 is divisible by 3, we must have $a \equiv 1$ (mod 3), otherwise one of $a^{n+1}$ and $(a+1)^n$ is a multiple of 3 and the other is not, so their difference cannot be divisible by 3. Now $a^{n+1} \equiv 1$ (mod 3), so we must have $(a+1)^n \equiv 1$ (mod 3), which forces $n$ to be even, and in particular at least 2.

If $a$ is even, then $a^{n+1} - (a+1)^n \equiv -(a+1)^n$ (mod 4). Since $n$ is even, $-(a+1)^n \equiv -1$ (mod 4). Since $2001 \equiv 1$ (mod 4), this is impossible. Thus $a$ is odd, and so must divide $1001 = 7 \times 11 \times 13$. Moreover, $a^{n+1} - (a+1)^n \equiv a$ (mod 4), so $a \equiv 1$ (mod 4).

Of the divisors of $7 \times 11 \times 13$, those congruent to 1 mod 3 are precisely those not divisible by 11 (since 7 and 13 are both congruent to 1 mod 3). Thus $a$ divides $7 \times 13$.

Now $a \equiv 1$ (mod 4) is only possible if $a$ divides 13.