SOLUTION: Determine all natural numbers n such that for all positive divisors d of n, d+1 is also a divisor of n+1.

n cannot be even because then it would have divisor d=1 and d+1 would be 2,
which could not be a divisor of n+1 for n+1 would be odd.  So we know that n
must be odd.

Every odd natural number n has positive divisor d=1, and d+1=2 will always 
be a divisor of the next integer n+1, for n+1 will be even. 

Every odd integer n also has positive divisor d=n (itself), and d+1
(which equals n+1) will always be a divisor of n+1 (itself).

We ask: Can n have divisors other that 1 and n itself?

Let's find out by dividing n+1 by d+1 by long division, assuming d is not 1.

      n/d_______
d + 1) n  +   1
       n  +  n/d
            1-n/d = remainder = 0

The remainder here must be 0, so





So the answer is no, because d=n tells us that every divisor d other than 1 of n
must be equal to then natural number n itself.  So n cannot have any divisors
other than itself and 1.  The only natural numbers n that only have divisors
1 and themselves are 1 and the odd primes.

Answer: n is either 1 or an odd prime number.

Edwin