Peter wrote: Prove that there are unique positive integers $a$ and $n$ such that \[a^{n+1}-(a+1)^{n}= 2001.\] $-(a+1)^{n}\equiv 2001 \mod a$ $-1 \equiv 2001 \mod a$ $0 \equiv 2002 \mod a$ a=1, 2, 7, 11, 13, 154, 182, 286, 1001, 2002 Don't all of these work for a?

All of them work for your equation, but it's not because they satisfy that mod a equation that they'll satisfy the original problem... The only thing you now know is that $a\in\{1, 2, 7, 11, 13, 154, 182, 286, 1001, 2002 \}$, now you have to prove that exactly one of those $a$'s has exactly one $n$ for which the equation is true.

As shown earlier a|2002. This means that a can only be 1, 2, 7, 11, 13, 14, 22, 26, 77, 91, 143, 154, 182, 286, 1001, 2002. Mod a+1 gives us -1^(n+1)=2001 mod a+1 so either a+1|2002 or a+1|2000 If a|2002 and a+1|2002, we have that a=1 or a=13. a=13 works (n=2). Our other case is a+1|2000. If this is true, then the only factors of a+1 are 2 and 5. Checking our factors above (and adding 1), we see that none are powers of only 2 and 5. So, our only solution is $ 13^{3}-14^{2}$

Hi. Could you explain why Mod a+1 gives us -1^(n+1)=2001 mod a+1? I do not seem to be able to get this part, maybe I do not get it. Mind explaining? thanks.

We are given $ a^{n+1} - (a+1)^n = 2001$, and you expressed no surprise that then $ 2001 \equiv -1 \pmod{a}$. Similarly, $ 2001 \equiv ((a+1)-1)^{n+1} - (a+1)^n \equiv (-1)^{n+1} \pmod{(a+1)}$.

Ah... I see it now. Haha. Really need to practice manipulating delicately, eh. Thanks for the explanation.

A quick way to prove that $a|2002$ is to expand the LHS, move 2001 to the LHS, and use the Rational Root Theorem.

works mod 3 gives n even, so only the case a+1|2002.

$ a | 2002 $