Is the number 100001 in base b always composite?

Question

In base 2, 100001 = 33, not prime
In base 3, 100001 = 244, not prime
In base 4, 100001 = 1025, not prime

...?

No, base 1 doesn't count. b ≥ 2

Also, b is an integer -- base 36^(.2) doesn't count either. Have I closed all the loopholes? :)

Answer 1

The answer is "yes".
In base b, 100001 = b^5 + 1,
which is always divisible by b+1 for any b. In fact,
b^5 + 1 = (b+1)(b^4 -b³ + b² -b +1).

Answer 2

100001 in base n = n^5 + 1, so let's look at n^5 + 1.

If n is odd, the n^5 is odd because odd * odd = odd (implying that odd*odd*odd*odd*odd = odd), which means that n^5 + 1 is even --> if n is odd then 100001 base n is composite.

Obviously, if n = 2 then n^5 + 1 = 33 = composite, as you noted.

If n is even, then n=2p. n^5+1 = 32p^5+1 and (32p^5+1)/(2p+1) = (16p^4 - 8p^3 + 4p^2 - 2p + 1), so n^5+1 is composite.

Therefore, n^5 + 1 is always composite.