Using pumping lemma show that the language L, consisting of all string of 1’s whose length is a prime is not a regular language.

If we show that the language L_pr consisting of all strings of 1’s whose length is a prime is not a regular language. Then there would be a constant n satisfying the condition of the pumping lemma.
Consider some prime p ≥ n + 2, there must be such a p, since there are an infinity of primes,
Let, w = 1^p
By the pumping lemma, we can break w = xyz such that y ≠ ϵ and | xy | ≤ n
Let, | y | = m, then | xz | = p – m
Now consider the string xy^(p-m) z, which must be in L_pr by the pumping lemma, if L_pr really is regular. However,
| xy^(p-m) z | = | xz | + (p - m) | y | = p – m + (p – m)m = (m + 1) (p – m)
It looks like | xy^(p-m) z | is not a prime, since it has two factors m + 1 and p – m. However, we must check that neither of these factors are 1, since then (m + 1) (P – m) might be a prime after all. But m + 1 > 1, since y ≠ ϵ tells us m ≥ 1. Also, p – m >1, since p ≥ n + 2 was chosen and m ≤ 1. Also, p – m > 1, since p ≥ n + 2 was chosen and m ≤ n since.
m = | y | ≤ | xy | ≤ n
Thus, p – m ≥ 2.
Again, we have started by assuming the language in question was regular or and we derived a contradiction by showing that some string not in the language was required by the pumping lemma to be in the language. Thus, we conclude that L_pr is not a regular language.