Consider the following sequence:

10001, 100010001, 1000100010001, 10001000100010001, 100010001000100010001,....

Show that none of them are prime numbers.

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 "tetra-productial 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)

10^4-1 is, of course, always equal to 9999, and 10^2-1 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 right-hand 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 2003-05-23 20:23:11