Putnam 2001 
Topic: Algebra Number Theory

Problem: Find unique positive integers $a$, $n$ such that $a^{n+1}-(a+1)^n=2001$ and give their product as the final answer.

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 $a \equiv 1 \pmod{3}$, 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 $a^{n+1} \equiv 1 \pmod{3}$, so we must have $(a+1)^n \equiv 1 \pmod{3}$, which forces $n$ to be even, and in particular at least 2.  If $a$ is even, then $a^{n+1} - (a+1)^n \equiv -(a+1)^n \pmod{4}$. Since $n$ is even, $-(a+1)^n \equiv -1 \pmod{4}$. Since $2001 \equiv 1 \pmod{4}$, this is impossible. Thus $a$ is odd, and so must divide $1001 = 7 \times 11 \times 13$. Moreover, $a^{n+1} - (a+1)^n \equiv a \pmod{4}$, so $a \equiv 1 \pmod{4}$.  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 $a \equiv 1 \pmod{4}$ is only possible if $a$ divides $13$.  We cannot have $a=1$, since $1 - 2^n \neq 2001$ for any $n$. Thus the only possibility is $a = 13$. One easily checks that $a=13, n=2$ is a solution; all that remains is to check that no other $n$ works. In fact, if $n > 2$, then $13^{n+1} \equiv 2001 \equiv 1 \pmod{8}$. But $13^{n+1} \equiv 13 \pmod{8}$ since $n$ is even, contradiction. Thus $a=13, n=2$ are the unique values with the final solution being \boxed{26}.