2023/007) FInd integer n such that $n^2+ 19n = n!$

 n = 0 is not a solution so $n > 0$.

deviding both sides by n we get

$n + 19 = (n-1)!$

putting n = x+ 1 we get $x+20 = x!$

for this to be valud x is a factor of 20

so we check with 1 LHS = 21 and RHS = 1

x = 2 LHS = 22 RHS = 2 no

x = 4 LHS = 24 LHS = 24 so x = 4 is a solution

x = 5 LHS = 25 RHS = 125 not a solution

if we take x larger RHS grows larger as compared to LHS so no solution

so solution x = 4 or n = 5.