Find $a,n\in \mathbb N^{+}:a!+\dfrac{n!}{a!}=x^2,x\in \mathbb N$

Find $a,n\in \mathbb N^{+}:a!+\dfrac{n!}{a!}=x^2,x\in \mathbb N.$

I find $\{n,a\}=\{4,1\}\{4,4\}\{5,1\}\{5,5\}\{7,1\}\{7,7\}\{20,11\}.$ (These are all if $n<300$.)

$11!+\dfrac{20!}{11!}=1\times2\times3\times4\times5\times6\times7\times8\times9\times10\times11+12\times13\times14\times15\times16\times17\times18\times19\times20=246960^2.$

If we assume abc conjecture, will the solution be finite?

All inequalities here hold for $n$ being suffficiently large. It is trivial to show that $$\frac{n!}{(\frac{n}{2})!} < a! + \frac{n!}{a!} < 2n!$$

Now substitute $x^2$ and take log. You will obtain $$\frac{\ln n! - \ln \frac{n}{2}!}{2} < \ln x < \frac{\ln n! + \ln 2}{2}$$

Now calculate the quality from abc-hypothesis: $q(a!, \frac{n!}{a!}, x^2) = $ $$\frac{2 \ln x}{\ln n\# rad x} > \frac{2 \ln x}{\ln n\# + rad x} > \frac{2 \ln n! -2 \ln \frac{n}{2}!}{\ln n\# + \ln n + \ln 2} > \frac{2 \ln n! -2 \ln \frac{n}{2}!}{\frac{3}{2} \ln n! + ln 2} > 1+ \epsilon$$

The last comes from fact that $\frac{\ln \frac{n}{2}!}{\ln n!} \to 0$. Now we can apply abc-hypothesis.