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.
Let f(n) = 8n^4+15n^3 - 4n^2 +15n+8
Then,
f(1)=42, f(2)= 270, f(3)=1070, f(4)= 3012, f(5)= 6858, f(6) = 13562. f(7) = 24270, f(8)= 40320, f(9) = 63242, f(10) = 94758, f(11) = 136792
Comparing the foregoing with the values of n!, we observe that:
n! < f(n) for n=1,2,3,4,5,6,7
n! = f(n) for n=8
n!> f(n) for n=9, 10, 11
Accordingly, n=8 is a solution.
and it leads us to conjecture that n! > f(n) for every n>=9
Now, substituting n=m+8, we have:
f(m+8) = 8m^4+ 271m^3+ 3428m^2 +19215m +40320
By the given conditions:
(m+8)!=f(m+8l
Then, (m+9)!
= (m+9)*(m+8)!
= (m+9)*f(m+8)
=8m^5+ 343m^4+ 5867m^3+ 50067m^2+213255m+362880
> 8m^4 +303m^3+4289m^2+26916m+63242
=f(m+9), whenever m>=0
Therefore, (n+1)! > f(n+1), whenever n>=8
or, n! > f(n) whenever n>=9
Accordingly, no solution exists whenever n>=9
Consequently, n=8 is the only possible solution to the given puzzle.
** I call this "semi-analytic" since I solicited the aid of online solvers for computing f(m+8), f(m+9) and evaluating th product of large polynomials.
Edited on July 2, 2022, 7:25 am
Semi-Analytic Puzzle Solution: Part 1
