Is $100x^2+20x$ a factorial?

Question

In this equation , $$100x^2+20x =m! $$where $x$ and $m$ are integers arethere infinite results of this to be true. It may be easier to think that when $m! = 1-9$ can be solved easily $$m!>9 $$ so that $$m!=1×2×3×4×5×6×7×8×9×10\times\ldots×m!$$ Given the equation $$100x^2+20x=m!$$ $$20(5x^2+x)=m!$$ $$5x^2+x=\frac{m!}{20}$$ And $\dfrac{m!}{20} \bmod 5 =0$

Answer 1

Take $y = 10x + 1$, the equation becomes $y^2 = m! + 1$ and reduce to a special case of an open problem ( Brocard's problem ).

As of this moment, only three solutions for $(m,y)$ is known. $(4,5)$, $(5,11)$ and $(7,71)$. This leads to two solutions of current problem $(m,x) = (5,1)$ and $(7,7)$.

Look at the wiki page and references there for more info.