Show that for any prime p there exists a nonnegative integer N such that:
2^N+3^N+6^N-1 is a multiple of p.
Equivalently, for some integer multiple, k, k=(2^n+3^n+6^n-1)/p.
Checking 'small' values:
n | 2^n + 3^n + 6^n - 1
1 | 10 Divisible by 2 and 5
2 | 48 Divisible by 2 and 3
3 | 250 Divisible by 2 and 5
4 | 1392 Divisible by 2 and 3
5 | 8050 Divisible by 2, 5 and 7
6 | 47448 Divisible by 2 and 3
7 | 282250 Divisible by 2 and 5
8 | 1686432 Divisible by 2, 3 and 11
9 | 10097890 Divisible by 2, 5 and 11
10 | 60526248 Divisible by 2 and 3
11 | 362976250 Divisible by 2,5, 7 and 13
So 2,3,5,7,11,and 13 are already accounted for.
Note though that there is a pattern: when n=3, 5 divides the total, when n=5, 7 divides it,... when n=11, 13 divides it.
This leads to the conjecture: for some prime (2n+1), and exponent (2n-1): (2n+1)k=(2^(2n-1)+3^(2n-1)+6^(2n-1)-1). Call RHS 'the expression'
Checking: this is true for k,n x=(2n+1):
{k == 50, n == 2, x == 5},
{k == 1150, n == 3, x == 7},
{k == 917990, n == 5, x == 11},
{k == 27921250, n == 6, x == 13},
{k == 27658786250, n == 8, x == 17},
{k == 890883616630, n == 9, x == 19}}
For n=4, k=282250/9 and x=9, but 9 is not prime
For n=7, k = 2612459306/3, and x=15, but 15 is not prime.
When dealing with exponents that are a bit more or less than a prime, there is a clear motivation towards Fermat, a^p==a(modp)
For a^(p-2) the rules are more complex; by observation of small values for a={2,3,6}:
Where a=2, a^(p-2) = (p+1)/2 modp
Where a=3, a^(p-2) = (p+1)/3 modp for primes of form (6k-1), and (2p+1)/3 modp for primes of form (6k+1)
Where a=6, a^(p-2) = (p+1)/6 modp for primes of form (6k-1), and (5p+1)/6 modp for primes of form (6k+1)
Now for some prime p, of form (6k-1) we have (p+1)/2+(p+1)/3+(p+1)/6 = p+1 for the powers-1 = 0, modp; p divides the expression.
And for some prime p, of form (6k+1) we have (p+1)/2+(2p+1)/3+(5p+1)/6 =2p+1 for the powers-1 = 0, modp; p divides the expression.
And we are done.
Solution