First some kind of sequential representation of the numbers might be in order. Each of these numbers can be expressed as:
1 + 10^4 + 10^8 + ... + 10^4i
For any of these numbers, it can be said:
10^(4i+4) 1 = (10^4  1)(1 + 10^4 + 10^8 + ... + 10^4i)
for example, if i=3:
999999999999 = 9999 * 100010001
(Remember the solution to the "tetraproductial numbers" problem
Similary, the same could be said for numbers of the form:
1 + 10^2 + 10^4 + 10^6 + ... + 10^2i
where:
10^(2i+2) 1 = (10^2  1)(1 + 10^2 + 10^4 + ... + 10^2i)
for example:
99999999 = 99 * 1010101
Finally, we can say that:
10^(4i+4) 1 = (10^(2i+2) 1)(10^(2i_2) +1)
for example, if i=2:
99999999 = 9999 * 10001
Going back to the original equations, we have:
10^(4i+4) 1 = (10^4  1)(1 + 10^4 + 10^8 + ... + 10^4i)
10^(2i+2) 1 = (10^2  1)(1 + 10^2 + 10^4 + ... + 10^2i)
but since
10^(4i+4) 1 = (10^(2i+2) 1)(10^(2i_2) +1)
we can say:
10^(4i+4) 1 = (10^2  1)(1 + 10^2 + 10^4 + ... + 10^2i)(10^(2i_2) +1)
and thus:
(10^4  1)(1 + 10^4 + 10^8 + ... + 10^4i) = (10^2  1)(1 + 10^2 + 10^4 + ... + 10^2i)(10^(2i_2) +1)
10^41 is, of course, always equal to 9999, and 10^21 is always 99. So, putting these onto the same side of the equations (9999/99=101):
101(1 + 10^4 + 10^8 + ... + 10^4i) = (1 + 10^2 + 10^4 + ... + 10^2i)(10^(2i_2) +1)
Since 101 is a prime number, one [or both] of the factors on the righthand side of the equation must have 101 as a fator. If i>1, then for whichever number is divisible by 101, the quotient will exceed 1, meaning (1 + 10^4 + 10^8 + ... + 10^4i) is expressible as the product of two factors., each greater than 1. If k=1, we have the number 10^4 1 = 10001, which is composite (73*137).

Posted by DJ
on 20030523 20:23:11 