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## Main Question or Discussion Point

"11 and 101 are primes, whereas 1001 is not (7*11*13). Find the next smallest prime of the form 100...01, and show that it is the next"

I'm fairly sure that there no primes of the form 10^k + 1 above 101, but I can't seem to find a complete proof.

Consider the general case of factorising x^n + 1 = 0 (the above problem is the special case at x = 10)

If n were odd, then there would be a root of x^n + 1 = 0 at x = -1, and hence (x + 1) is a factor of (x^n + 1) where n is odd. (transfers to 11 dividing 11, 1001, 100001, etc)

If n = 2 (mod 4), then there is a solution at x = i to x^n + 1 = 0, so (x^2 + 1) is a factor (transferring to 101 dividing 101, 1000001, 10000000001)

This does not extend to when n = 0 (mod 4), as x^4 + 1 cannot be factorised, nor can x^8 + 1, etc. However, 10^4 + 1, 10^8 + 1, etc, are all composite numbers, up to 1300 (according to various sources on the internet).

Am I going down completely the wrong path, or what?

I'm fairly sure that there no primes of the form 10^k + 1 above 101, but I can't seem to find a complete proof.

Consider the general case of factorising x^n + 1 = 0 (the above problem is the special case at x = 10)

If n were odd, then there would be a root of x^n + 1 = 0 at x = -1, and hence (x + 1) is a factor of (x^n + 1) where n is odd. (transfers to 11 dividing 11, 1001, 100001, etc)

If n = 2 (mod 4), then there is a solution at x = i to x^n + 1 = 0, so (x^2 + 1) is a factor (transferring to 101 dividing 101, 1000001, 10000000001)

This does not extend to when n = 0 (mod 4), as x^4 + 1 cannot be factorised, nor can x^8 + 1, etc. However, 10^4 + 1, 10^8 + 1, etc, are all composite numbers, up to 1300 (according to various sources on the internet).

Am I going down completely the wrong path, or what?