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
Solution To The Problem
