If (2^32 + 1) is completely divisible by a whole number, which of the following numbers is completely divisible by this number?  Pick one
A) (2^96+1) B) (7 × 223 ) C) (2^16 - 1) D) (2^15 + 1) E) (2^17 + 1)
Explanation:   Let   2   32   =   x   232=x.   Then   (   2   32   +   1   )   =   (   x   +   1   )   (232+1)=(x+1)   (2^96+1)ssume that   (   x   +   1   )   (x+1) is completely divisible by a whole number, N   (   2   96   +   1   )   =   (   2   32   )   3   +   1   =   (   x   3   +   1   )   =   (   x   +   1   )   (   x   2   −   x   +   1   )   (296+1)=(232)3+1=(x3+1)=(x+1)(x2−x+1)   if   (   x   +   1   )   (x+1) is completely divisible by N,   (   x   +   1   )   (   x   2   −   x   +   1   )   (x+1)(x2−x+1) will also be divisible by N   Hence   (   2   96   +   1   )   (2^96+1) is completely divisible N   (2^96+1))
<result>A</result>
If (2^32 + 1) is completely divisible by a whole number, which of the following numbers is completely divisible by this number?