- 13

- 0

i dont think they do as they would have to be even with an even number of (odd*even) factor pairs.

And if you assume this number is in the form of 2p where p is odd and has only two (odd*even) facots pairs, it must have the factors (2*psqaured) and (p * 2p)

if is was slightly excessive;

x + 1 = 1 + 2 + psqaured + p + 2p

x = (p+1)(p+2

so another factor pair....