Integer solutions of the equation $a^{n+1}-(a+1)^n = 2001 $ I am doing number theory and I came across that question $a^{n+1}-(a+1)^n = 2001 $. Find the integer solutions and show that they are the only solution.
 I really tried hard but i am nowhere near solution.
 A: I'm assuming $a,n$ are positive integers.
First of all, $a^{n+1}-(a+1)^n \equiv -1 \equiv 2001 \;\; \text{mod} \; a ,$ so $a$ divides $2002=2 \times 7 \times 11 \times 13. $ Because $3|2001$ then $a^{n+1} \equiv (a+1)^n \;\; \text{mod} \; 3.$ You can easily eliminate $a \equiv 0,-1 \;\; \text{mod} \; 3,$ therefore, $a \equiv 1 \;\; \text{mod} \; 3 \;(\star)$ and $n$ has to be even.
Case 1. If $a$ is even, then $ a^{n+1}-(a+1)^n \equiv -1 \;\; \text{mod} \; 4$ but $2001 \equiv 1 \;\; \text{mod} \; 4$ and $2 \not \equiv 0 \;\; \text{mod} \; 4$ so there's no solution.
Case 2. If $a$ is odd then $a^{n+1}-(a+1)^n \equiv a \;\; \text{mod}\; 4$ as $a^n \equiv 1 \;\; \text{mod} \; 4.$ Thus, $a \equiv 2001 \equiv 1 \;\; \text{mod} \; 4 \; (\star \star).$  From $(\star), (\star \star)$ the only possibility is $a=13.$
Now, it boils down to solving $13^{n+1}-14^n=2001.$ If $n>2$ then $14^n \equiv 0 \;\; \text{mod} \; 8$ while $13^{n+1} \equiv 5^{2k+1} \equiv 25^k. 5 \equiv 5 \;\; \text{mod} \; 8$ and $5 \not \equiv 1 \;\; \text{mod} \; 8.$ Hence, $n=2$ and $(a,n)=(13,2)$ is the only solution. 
A: I consider this as a fairly easy problem. Here is my approach-
Since RHS is divisible by $3$, the LHS is also divisible by $3$. But considering modulo $3$, $x$ must be congruent to $1$ mod $3$. Hence, considering mod $x$, i got $x$ $|$ $2002$.
Then considering mod ${x+1}$, i again got ${x+1}$$|$$2002$.
Now when i found the factors of $2002$, i got only $1$,$2$, and $13$,$14$ are the consecutive divisors of $2002$. But $x$$=$$1$ did not give solutions. So only possible case for $x$ was $13$. Now, i got that only $n$$=$$2$ satisfies. Hence there was only one pair $(13,2)$.
