An algebra problem by Ankit Vijay
For an integer , let denote the number of positive integers that divide . Suppose that m is a positive integer and are real numbers such that for all . Find the smallest possible value of
The answer is 28.
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1 solution
We want to find the minimal degree m such that there exists a polynomial p(x) of degree m, that when multiplied to the generating function
F ( x ) = n = 0 ∑ ∞ f 9 ( n ) x n = n = 1 ∑ 9 1 − x n 1 the denominator of F ( x ) is ta most ∏ n = 1 9 Φ n ( x ) where Φ n ( x ) denotes the nth cyclotomic polynomial By explicitly writing out the numerator as a product of cyclotomics, we can see that it does not have a root at any of the nth primitive roots of unity, where n is an integer between 1 and 9 Since m is at least the degree of the denominator, the minimum value is just φ ( 1 ) + φ ( 2 ) + . . . + φ ( 9 ) = 8 + ( 9 − 1 ) + ( 5 − 1 ) + ( 7 − 1 ) + 2 = 2 8 .