Derive Likelihood Ratio test for testing $H_0: \mu=\sigma^2$, $H_1: \mu \not= \sigma^2$.

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Let $X_1, . . . , X_n$ be independent $N(µ, σ^2)$ random variables. Derive the LRT for testing $H_0: \mu=\sigma^2$, $H_1: \mu \not= \sigma^2$.

As per usual I found MLE of $\theta \in \Theta$ and $\theta \in \Theta_0$ to calculate:

$$\Lambda=\frac{\sup_{\theta \in \Theta_0} L(\theta)}{\sup_{\theta \in \Theta}L(\theta)}=\frac{L(a,a)}{L(\bar{X},S^2)}$$ Where $a=\sqrt{\bar{X^2}+1/4}-1/2$ and $S^2=(1/n)\sum (X_i-\bar{X})^2$.

$L(a,a)=(2\pi a)^{-n/2}exp(\frac{-1}{2a}\sum (X_i-a)^2)$ $L(\bar{X},S^2)=(2\pi S^2)^{-n/2}exp(\frac{-1}{2S^2}\sum (X_i-\bar{X})^2)=(2\pi S^2)^{-n/2}exp(\frac{-1}{2S^2}nS^2)=(2\pi S^2)^{-n/2}exp(\frac{-n}{2})$

So I'm getting $$\Lambda=(\frac{eS^2}{a})^{n/2}\exp\{{\frac{-1}{2a}\sum(X_i-a)^2}\}$$

And here I'm stuck. Do I (can I even) simplify this more or can I just shove this into the LRT and move on? In the examples I'm given; they typically find the distribution of the statistic or find an expression in terms of some known distribution.

(1) Did I make a mistake?

(2) Can I simplify this further? In other words; how do I finish the problem that isn't just ''The LRT is $\chi[\Lambda \leq c]$''