If (2^32 + 1 ) is completely divisible by whole number.. which of the number will completely divisible by this number ??
A) ( 2^16 + 1)
B) ( 2^16 -1)
C) 7 * 2^33
D) (2^96 + 1)

• Let 2^32 = x. Then, (2^32 + 1) = (x + 1).
  Let (x + 1) be completely divisible by the natural number N. Then,
  (2^96 + 1) = [(2^32)^3 + 1] = (x^3 + 1) = (x + 1)(x^2 - x + 1), which is completely divisible by N, since (x + 1) is divisible by N.
• 9 years agoHelpfull: Yes(18) No(0)
• a^n+b^n is divisible ny a+b when n is a prime number
  so ((2^16)^2+1)is divisible by 2^16+1
• 9 years agoHelpfull: Yes(0) No(5)