If $L$ is a regular language then so is $\sqrt{L}=\{w:ww\in L\}$ I am interested in proving that $\sqrt{L}=\{w:ww\in L\}$ is regular if $L$ is regular but I don't seem to be getting anywhere. If possible I was hoping for a hint to get me going in the right direction. Thank you for your help.

My idea for demonstrating regularity of the square root language was to try to consider two machines running in tandem. One of them accepts the original language $L$ and the other runs the same machine backwards (I expect that this will be an NFA). I then wanted to accept words that met in the middle (i.e ${(q,q):q∈Q}$  where $Q$  are the states of the DFA accepting $L$ ). However, I do not think this will work.
Here is how to implement your solution. Let $A = \langle Q, q_0, F, \delta \rangle$ be a DFA for $L$. We will construct an NFA $A' = \langle Q', q'_0, F', \delta' \rangle$ as follows:


$Q' = \{q'_0\} \cup Q^3$. The state $(q_1,q_2,q_3)$ means that we have guessed that when $A$ finishes reading the first copy of $w$, it will be in state $q_1$; the first copy of $A$, started at $q_0$, is at state $q_2$; and the second copy of $A$, started at $q_1$, is at state $q_3$.
$F' = \{(q_1,q_1,q_2) : q_1 \in Q, q_2 \in F\}$. Thus we accept if the first copy of $A$ is in the guessed state, and the second copy of $A$ is at an accepting state.
$\delta'(q'_0,\epsilon) = \{(q,q_0,q) : q \in Q\}$. This initializes the simulation of the two copies of $A$.
$\delta'((q_1,q_2,q_3),a) = \{(q_1,\delta(q_2,a),\delta(q_3,a))\}$. This simulates both copies of $A$, while keeping the guessed state.


We leave the reader the formal proof that $L(A') = \sqrt{L(A)}$.



Here is another solution, which creates a DFA. We now run $|Q|$ copies of $A$ in parallel, starting at each state of $A$:


$Q' = Q^Q$.
$q'_0 = q \mapsto q$, the identity function.
$\delta'(f,a) = q \mapsto \delta(q,a)$. 
$F' = \{ f \in Q' : f(f(q_0)) \in F \}$.


What is the meaning of the condition $f(f(q_0)) \in F$? After reading a word $w$, the automaton $A'$ is in a state $f$ given by $f(q) = \delta(q,w)$. Thus $f(f(q_0)) = \delta(\delta(q_0,w),w) = \delta(q_0,w^2)$.