2^2(Mod 5) = -1, giving 2^58 (Mod 5) = -1, so that :

(2^58+1)(Mod 5) = 0…..(#)

Accordingly, (2^58+ 1)/5 must be an integer.

Now, we observe that:

2^58 + 1

= (2^29 + 1)^2 – (2^15)^2

= (2^29 + 2^15 + 1)(2^29 – 2^15+1)

Now, 2^29 +/- 2^15 +1 > 2^15(2^14+/- 1) + 1 > 2^15> 5

…….(##)

Since both the factors of 2^58+1 is greater than 5, it follows from (#) that at least one of the factors in (##) must be evenly divisible by 5. Hence:

(2^58+1)/5= xy, where both x and y must be positive integers, each greater than 1.

*Consequently, (2^58 + 1)/5 must be a composite whole number.*

*Edited on ***April 11, 2007, 1:07 am**