Let p and q be two different prime numbers greater than 3.Prove that if their difference is 2^n, then for any two integers n and m,the number S=p^(3m+1)+q^(2m+1) is divisible by 3.
Let p and q be two different prime numbers greater than 3.
Prove that if their difference is 2^{n}, then for any two integers n and m,
the number S = p^{(3m+1)} + q^{(2m+1)} is divisible by 3.
Let p and q be two different prime numbers greater than 3.
Prove that if their difference is a positive power of 2, then for any two integers n and m, the number S=p^(2m+1)+q^(2n+1) is divisible by 3.
Explanation:
If a prime is larger than 3, then it is of the form (6k+1) or (6k1). If a prime of the first form is raised to any odd power, then it will be worth 1, mod 3, and if a prime of the second form is raised to any odd power, then it will be worth 2, mod 3. Hence their sum will be divisible by 3.
The power of two guarantees that one of each type of prime will be selected since if both are the same type their difference will be 6k6l =6(kl) which has 3 as a divisor.
Lastly, while it is true that a (6k+1) prime will always be worth 1 mod3 if raised to any power, including an even power, a (6k1) prime will also be worth 1 mod 3 if raised to an even power, which will cause the puzzle not to work. So it's better to stipulate that both powers must be odd.

Posted by broll
on 20121203 07:14:18 