a=13, n=2 is the only solution in positive integers to the given equation.
EXPLANATION:
At most one of a, a+1 is a multiple of 3.
If exactly one of a, a+1 is a multiple of 3, then a^(n+1)(a+1)^n
would not be a multiple of 3. This is a contradiction,
since 2001 is a multiple of 3.
Thus, neither of a, a+1 is a multiple of 3, hence a ==1(mod 3)
If n is odd, then:
a^(n+1)  (a+1)^n =2001
=> a^(n+1)(a+1)^n 2001(mod 3)
=> 1^(n+1)2^n ==0 (mod3)
This leads to a contradiction.
Hence, n is even
Then, we must have:
a^(n+1)(a+1)^n =2001
=> a^(n+1)  (a+1)^n == 2001 (mod (a+1))
=> (1)^(n+1)  0^n == 2001(mod a+1)
=>1 ==2001(mod a+1)
=> 2002 == O (mod a+1)
=> a+1  2002
As previously noted, we also have a 2002
Thus, both a and a+1 are divisors of 2002
Since gcd(a, a+1) =1,
It follows that:
a(a+1)  2002
Then, a(a+1)<=2002, so a<45.
From the prime factorisation of:
2002= (2)*(7)*(11)*(13)
We get that the only positive integer dovisors of 2002 which are less than 45 corresponds to: 1,2,7,11,13,14,22,26.
Of these potential values, only a=10000 and a=13 are such that a+1 is a divisor of 2002. But, if a=1, then a^(n+1)  (a+1)^n = 1  2^n This leads to a contradiction.
It remains to analyse the case of a=13
Suppose n>2, then:
13^(n+1)14^n =2001
=> 13^(n+1)14^n ==2001 (mod8)
=>13^(n+1)== 1(mod 8)
=>(13^n)*(13)== 1(mod 8)
=> (1)*(5)==1 (mod 8)
This leads to a contradiction.
Since n is even, the only remaining possibility is n=2
By direct evaluation, it can be verified that: 13^314^2=2001
Therefore, a=13, n=2 is the only solution in positive integers to the given equation.
