String.prototype.isRepeated

Python (24)

lambda s:s in(s+s)[1:-1]

Checks if the string is a substring of itself concatenated twice, eliminating the first and last characters to avoid trivial matches. If it is, it must be a nontrivial cyclic permutation of itself, and thus the sum of repeated segments.

Regex (ECMAScript flavour), 11 bytes

Sounds like a job for regex!

^([^]+)\1+$

Test it here.

I've chosen ECMAScript, because it's the only flavour (I know) in which [^] matches any character. In all others, I'd either need a flag to change the behaviour of . or use [\s\S] which is three characters longer.

Depending on how we're counting the flag, that could of course be a byte shorter. E.g. if we're counting pattern + flags (e.g. ignoring delimiters), the PCRE/Perl equivalent would be

/^(.+)\1+$/s

Which is 10 bytes, ignoring the delimiters.

Test it here.

This matches only strings which consist of at least two repetitions of some substring.

Here is a full 26-byte ES6 function, but I maintain that regular expression submissions are generally valid:

f=s->/^([^]+)\1+$/.test(s)

Pyth, 9

/:+zz1_1z

Or

}z:+zz1_1

These are both close translations of @xnor's python answer, except that they take input from STDIN and print it. The first is equivalent to:

z = input()
print((z+z)[1:-1].count(z))

0 for False, 1 for True.

The second line is equivalent to:

z = input()
print(z in (z+z)[1:-1])

False for False, True for True.

Pyth's official compiler had a bug related to the second one, which I just patched, so the first is my official submission.