Prove that there are no prime numbers in the infinite sequence of integers 10001, 100010001, 1000100010001, ..

Question

i'm sorry i'm stupid....
i need help
thank you

i can't understand the explanations in other sites
please use simple words as much as possible
thanks a lot!

Answer 1

S(k) = 1 + 10^4 + 10^8 + . . . + 10^(4k)
for some positive integer k.

Note the three following identities:

10^(4k+ 4) - 1 = (10^4 - 1)(1 + 10^4 + 10^8 + . . . + 10^(4k)),

10^(2k + 2) - 1 = (10^2 - 1)(1 + 10^2 + 10^4 + . . . + 10^(2k)),

10^(4k+4) - 1 = (10^(2k+2) - 1)( 10^(2k+2) + 1).

Combining these,

(10^4 - 1)(1 + 10^4 + 10^8 + . . . + 10^(4k)) = 10^(4k+4) - 1
= (10^(2k+2) - 1)( 10^(2k+2) + 1).
= (10^2 - 1)(1 + 10^2 + 10^4 + . . . + 10^(2k)) (10^(2k+2) + 1)

Since (10^4 - 1)/(10^2 -1) = 100 + 1 = 101, we have

(1 + 10^4 + 10^8 + . . . + 10^(4k))*101
= (1 + 10^2 + 10^4 + . . . + 10^(2k)) ( 10^(2k + 2) + 1)

Since 101 is a prime number, at least one of the two factors on the right hand side is divisible by 101. If k > 1, then for whichever of them is divisible by 101, the quotient will exceed 1; hence 1 + 10^4 + 10^8 + . . . + 10^(4k) is expressible as a product of two factors, each greater than 1. When k=1 we have 10001 as a factor which is composite (73*137).

Answer 2

Well, I don't think anyone can say you are stupid, because this one isn't easy. After studying it for awhile, I think the secret may be the following factoring trick:

(1 + x + x^2 + x^3 + ... + x^k)(1 - x + x^2 - x^3 + ... + x^k) = 1 + x^2 + x^4 + ... + x^(2k) for any even integer k.

This gives you a factor anytime the number of 1s in the number is odd. For example 10001000100010001 has five 1s, which is equal to 10^16 + 10^12 + 10^8 + 10^4 + 1. To use the factorinng trick, set x to 100. The factoring is

(1 + 100 + 100^2 + 100^3 + 100^4)(1 - 100 + 100^2 - 100^3 + 100^4) = 1 + 10^4 + 10^8 + 10^12 + 10^16

So, 101010101 is a factor is 10001000100010001.

Now, the above factoring trick only works if k, i.e the highest exponent, is EVEN. That's important because the second factor, with the alternating + and - signs, has to finish with a + sign. That means so far we have only proven the numbers with an odd number of 1s are not prime. However, when the number is 1s even, then we actually have a simpler factorization: the number 10001 = 10^4 + 1 will always be a factor.

This leaves only one number to test, namely 10001. And, it turns out this number can be factored as 73 * 137.

Nope, definitely not an easly one...

Answer 3

a(0) = 1
a(1) = 10001 = 10000 + 1
a(2) = 100010001 = 100000000 + 10000 + 1

q = 10000 = 100^2
r = 100

a(n) = sum[0 to k] q^k = sum[0 to k] r^2k

a(n) = (r^2(n+1) - 1) / (r ^2 - 1) =
(r^(n+1) - 1)(r^(n+1) + 1) / 99 * 101

if a(n) is prime, then 99*101*p = (r^(n+1) - 1)(r^(n+1) + 1),
which is impossible, because if either of operands has
p as a factor, then the other one must be less or equal to
101*99 = 9999

Answer 4

Prime numbers are infinite. I know for sure that some discovered a 17 digit prime number and there is no formula for testing it other than common sense.