x^2=n!+1⇒ (x+1)(x-1)=n! where (x+1)/2 and (x-1)/2 are consecutive integers and have consecutive primes as factor ,let ,y and z (respectively) so it can be written y-1=z. Consider prime counting function π(z),π(2z-1) that count primes less than the variable or argument. It can be seen that f(z) is equals to the numbers of z-k when z-k is prime where 0<k<z+1.Same way , the total number of primes in between z+1 and 2z-1 is also the number of primes of the form z+k in between that interval. So, it can be written that, π(z)-w(z)= π(2z-1)- π(z)(ignoring the prime 2) where w(z) is the number of distinct prime divisor of z. so, 2π (z)- π(2z) = w(z).now form prime number theorem and an asymptotic formula for w(n), it can be written 2* ln2 * z =lnlnz *ln2z* lnz +f(z) (where f(z) is error term.) ⇒2* ln2 * z<<lnlnz *ln2z* lnz,which means after a certain z, there will be no intersection point. It is sort of outline,I would be happy if you find any flaw and post it here,thanks. I have a stack exchange link regarding the solution- real analysis - Number of solutions of asymptotic funtion's equation. - Mathematics Stack Exchange I explained the idea, but no response to that. Initially it was put on hold due to "unclear" tag. Any thoughts?