Solve for n (a positive integer) in each equation:
(1) n! = 8n^4 + 15n^3 - 4n^2 + 15n + 8
(2) n! = 9n^4 + 4n^3 + n^2 + 1344
Both analytic and computer solutions welcome.
As before letting f(n) = 9n^4+4n^3+n^2+1344, we have:
f(1)=1358, f(2)= 1524, f(3)=2190, f(4)= 3920, f(5) = 7494, f(6)= 13908, f(7)= 24370, f(8)= 40320, f(9)= 63390, f(10)= 95444, f(11)= 138558
Accordingly,
f(n) > n!, whenever n<=7
f(8)= 8!=40320
f(n) < n! for n=9, 10,11
Accordingly, n=8 is a solution and it leads us to conjecture that:
n!> f(n), whenever n>=9
Now, substituting n=m+8, we have:
f(m+8)= 9m^4+292m^3+3553m^2+19216m+40320
By the given conditions, we have:
(m+8)!=f(m+8)
=> (m+9)! = (m+9)*(m+8)!
=(m+9)*f(m+8)
= 9m^5++373m^4+6181m^3+51193m^2+213264m+362880
> 9n^4+328n^3+4483n^2+27234m+633390
=f(m+9)( by online solver computation)
Therefore, (n+1)! > f(n+1), whenever n>=8
Or, n! > f(n), whenever n>=9
Consequently, n=8 is the only possible solution to the given puzzle.
Edited on July 2, 2022, 10:02 am
Semi-Analytic Puzzle Solution: Part 2
