A New Year's Resolution: Pose Better Problems Than This One...
It's a New Year! Time for a problem that may appear to be rather odd and of little use, but quite challenging.
Consider the function f(n) = 3^{2n+5}+160n^{2}56n243....and note these patterns:
 When n = 0, f(0) = 0...which is divisible by 512
 When n = 1, f(1) = 2048...which is divisible by 512
 When n = 2, f(2) = 19968...which is divisible by 512
Some questions:
 Do you think f(n) is divisible for all whole number values of n? Explain.
 What happens if n = 1, 2, etc.? Elaborate.
 What happens if you try fractional values (nonintegers) of n? Elaborate.
Now the big step: Prove/Disprove that f(n) is divisible by 512 for all integer values of n.
Source: School Science and Mathematics, October 1956, Problem 2510.
Hint: This is a tough one...and if I try to give a hint, I perhaps spoil the problem.
Solution Commentary: The problem's solution by the proposer in 1956:
f(n+1)9f(n)= 8(160n^{2})+768n+2048
f(n+1)9f(n)= 512(4+n2n^{2})256n(n1)
f(n+1)9f(n)= 512K_{1}512K_{2}, since n(n1) is even
f(n+1)9f(n)= 512k, which implies that 512 divides f(n) for integer values of n
Now, I posed this problem to show how useless and frustrating mathematics can be sometimes. Who would ever dream up this solution from the problem, except perhaps for the poser of the problem? The "odd" thing is that at least five other people solved the problem when posed in 1956.
Also, by looking at the proof, do you see why things do not work out when n is not an integer?
