Does anyone know how to solve Diophantine equation: $${\sqrt{n}}^\sqrt{n}-11 =m!^2.$$
I tried to substitute $\sqrt{n}=k$ then equation becomes $$k^k-11=m!^2\\\implies k^k=m!^2+11=(m!-1)(m!+1)+12$$ which means suppose $m\geq2$ then $\gcd{(m!-1,m!+1)}=1$. Does this give any hint? I could think upto here only. Please help.
Let $$k^k-11 = (m!)^2.$$ If $k>=11$ then $$121\not|\, LHS,\quad 121\,|\, RHS,\quad LHS\not= RHS.$$ In the other hand, $k$ is odd and $k^k > 11$, so $$k\in\{3,5,7,9\}.$$ Note than: $$\sqrt{3^3-11} = 4 \not= m!,$$ $$\sqrt{5^5-11} = \sqrt{3114}\in(55,56),\quad \sqrt{5^5-11}\not\in\mathbb N,$$ $$\sqrt{7^7-11} = \sqrt{823532}\in(907,908),\quad \sqrt{7^7-11}\not\in\mathbb N,$$ $$\sqrt{9^9-11} = \sqrt{387420478}= \sqrt{11683^2-1}\in(11682,11683),\quad \sqrt{9^9-11}\not\in\mathbb N.$$
So the issue diophantine equation has not solutions.