[tex]\dfrac{x!-(x-3)!}{23}=1\\x!-(x-3)!=23\\(x-3)!((x-2)(x-1)x-1)=23\\(x-3)!((x^3-x^2-2x^2+2x)-1)=23\\(x-3)!((x^3-3x^2+2x)-1)=23[/tex]
23 is a prime number, therefore there are two possibilities:
[tex]\text{I.}\, (x-3)!=1 \wedge x^3-3x^2+2x-1=23[/tex]
or
[tex]\text{II.}\, (x-3)!=23 \wedge x^3-3x^2+2x-1=1[/tex]
[tex]\text{I.}\\(x-3)!=1\\x-3=0 \vee x-3=1\\x=3 \vee x=4[/tex]
Now, we check if any of these solutions is also a solution to the second equation:
[tex]3^3-3\cdot3^2+2\cdot3-1=23\\27-27+6-1-23=0\\ -18=0[/tex]
Therefore, 3 is not a solution.
[tex]4^3-3\cdot4^2+2\cdot4-1=23\\64-48+8-1-23=0\\0=0[/tex]
Therefore, 4 is a solution.
[tex]\text{II.}[/tex]
[tex](x-3)!=23[/tex]
We know that [tex]3!=6[/tex] and [tex]4!=24[/tex], therefore there isn't any [tex]n\in\mathbb{N}[/tex], for which [tex]n!=23[/tex], so there's no solution.
So, the only solution is [tex]x=4[/tex].
