This blast from the past came up as a random problem and looked interesting...

p^12 - 1 = 0 mod 13

p^4 = 1 mod 5 and so therefore is p^12 so p^12 - 1 = 0 mod 5

p^6 = 1 mod 7 and so therefore is p^12 so p^12 - 1 = 0 mod 7

p is odd, so p^2 = 1 mod 8, since (2n+1)^2 = 4n^2 + 4n + 1 = 4n(n+1) + 1 and one of n or n+1 must be even. Since p^2 = 1 mod 8, p^4 must be 1 mod 16 since (8n+1)^2 = 64n^2 + 16n + 1 = 16n(4n+1) + 1 if p^4 = 1 mod 16 then so does p^12

p is not 3 so p^3 = 1 mod 9 since (3n +- 1)^3 = 27n^3 + 27n^2 +9n + 1 = 9n(3n^2 + 3n + 1) + 1

Since p^3 = 1 mod 9, then so is p^12. 

Now consider the three primes P1, P2, P3. Rewrite the expression to distribute the 3 like:

(P1^12 - 1) + (P2^12 - 1) + (P3^12 - 1)

each of these three terms must be divisible by each of (5,7,13,9,16) from the above, which means it must be divisible by the product. 

Reorder the odd factors as (5*13)*(7*9) = 65*63 = (64 + 1)(64 - 1) = (2^6 + 1)(2^6 - 1) = (2^12 - 1). Now multiply by 16 = 2^4 to arrive at 2^16 - 2^4 which must divide the expression on the left.