Let S be the square of a prime number, such that S is greater than 41.

Prove that (S-41)(S+41) is evenly divisible by 240.

S = pē for some prime p. Since S > 41, p >=7 and so is odd.

S is therefore also odd, and more specifically is of the form 8k + 1 (as are all odd squares.)  Since 41 is also of the form 8k + 1, the term (S-41) is a multiple of 8. Since S and 41 are both odd, the term (S+41) is even. Collectively, the product is a multiple of 16 (8 from the first factor, 2 from the second.)

p is not 3 (it's >=7) and so p = 1 or -1 mod 3. S, then, = 1 mod 3. Since 41 = -1 mod 3, the factor (S+41) is a multiple of 3.

p is also not 5 and so p = +/- 1 or +/-2 mod 5. S, then, = +1 or -1 mod 5 respectively. Since 41 = 1 mod 5, if S = +1 mod 5 then the term (S-41) is a multiple of 5 and if S = -1 mod 5 then the term (S+41) is a multiple of 5. Either way, the product is a multiple of 5.

Since the product is a multiple of 16, of 3, and of 5, it is a multiple of 16*3*5 = 240.

Posted by Paul on 2011-04-25 20:29:17