Where $m$ and $x$ any real non-negative interger values. Prove that $$\frac{m!}{200}\neq50x^2+51x+13$$ Where $m!\geq20$
I understand that it may link into a trivial part of Brocard's problem. $$\frac{m!}{200}\neq50x^2+51x+13$$$$200(50x^2 + 51x + 13) + 1 = m!+1$$ $$200(50x^2 + 51x + 13) + 1 = (100x + 51)^2$$
The thing I'm working on is below Assuming that $\frac{m!}{200}=50x^2+51x+13$
Because $m\geq20$: $m!$ is divisible by $3$. Therefore $\frac{m!}{200}$ is divisible by $3$. Therefore $50x^2+51x+13$ is divisible by 3. You can apply this for any prime (apart from $5$). Therefore $50x^2+51x+13$ divides up to $m$ where $m$ doesn't divide by $5$ or $2$.
Expansion-I'm not sure if this is the correct way of going throught this Using similar methods from the one above :Because $m\geq20$: $m!$ is divisible by $13$. Therefore $\frac{m!}{200}$ is divisible by $13$.Therefore $\frac{m!}{200}-13$ is divisible by $13$. Therefore $50x^2+51x$ is divisible by $13$.
Because $50x^2+51x$ is divisible by 13 , $x$ must be divisble by $13$.
A possible continuation:
Let's call $13y=x$ because $x$ is divisble by 13 $\therefore$ $50((13y)^2)+51(13y)+13=\frac{m!}{200}$ This equals $8450y^2+663y+13=\frac{m!}{200}$. For ease I will use two fractions. $$13(650y^2+51y+1)=\frac{m!}{200}$$$$650y^2+51y+1=\frac{\frac{m!}{200}}{13}$$ $$650y^2+51y=\frac{\frac{m!}{200}}{13}-1$$ $$13(50y^2)+17(3y)=\frac{\frac{m!}{200}}{13}-1$$ $$$$ To be continued... if $m\geq26$ $m!$ is divisible by 169.