Prove that for n > 1 the polynomial xn/n! + xn-1/(n-1)! + ... + x/1! + 1 has no rational roots.
Solution
Let p be any prime. We show first that pk does not divide k! Take m so that pm ≤ k < pm+1. Then the highest power of p dividing k! is pr, where r = [k/p] + [k/p2] + [k/p3] + ... + [k/pm] ≤ k/p + k/p2 + ... + k/pm = k(1 - 1/pm)/(p - 1) < k.
Suppose there is a rational root x of the polynomial. Put x = a/b, where a and b are coprime integers. Then an + nan-1b + ... + (n-1)! a bn-1 + n! bn = 0. So an must be divisible by b. But a and b are coprime, so b must be 1, in other words, the root must be an integer.
Now take p to be a prime dividing n. Then p must divide an and hence a. But we have - n! = n! (an/n! + an-1/(n-1)! + ... + a1/1! ). Each term inside the parentheses is divisible by p, because pk divides ak but not k! So a higher power of p divides the rhs than the lhs. Contradiction.
© John Scholes
jscholes@kalva.demon.co.uk
28 Dec 2002
Last corrected/updated 28 Dec 02