Integer solutions of the equation $a^{n+1}-(a+1)^n = 2001 $

Question

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.

Answer 1

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.