I love prime numbers! 2

Notice that

$10101 ... 10101 = 1 + 10^{2} + 10^{4} + ... + 10^{2n}$

$= 10^{0} + 10^{2} + 10^{4} + ... + 10^{2n}$

$= \frac {1} {99} (10^{2n+2} - 1)$

Since $(2n+2) = 2(n+1)$ , by the difference of two squares this gives

$10101 ... 10101 = \frac{1}{99} (10^{n+1} - 1) (10^{n+1} + 1)$

When $n = 1$ :

$(10^{n+1} - 1) = 99$

$(10^{n+1} + 1) = 101$

When $n = 2$ :

$(10^{n+1} - 1) = 999$

$(10^{n+1} + 1) = 1001$

When $n = 3$ :

$(10^{n+1} - 1) = 9999$

$(10^{n+1} + 1) = 10001$

For $n > 1$ , the two factors $(10^{n+1} - 1) (10^{n+1} + 1)$ are both greater than $99$ , hence when $(10^{n+1} - 1)$ is divided by factors of $99$ , the two remain as non-trivial factors (i.e. factors that are not $1$ or the number itself) of the original number. Thus we can conclude that the number is not prime, except when $n=1$ .

When $n=1$ , the number is $101$ and that is a prime number.

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Other prime number problems:

I love prime numbers! 1

I love prime numbers! 3

I did it a bit unusual way 🙂🙂 Converting each sequence from binary to decimal 101=5 10101=21 1010101=85 101010101=341 We see that only 5 is prime. So answer is 1