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Sum of exponents of primes in factorisation

Source: Chinese TST 2 2013 Day 2 Q1

March 29, 2013
symmetrynumber theory proposednumber theory

Problem Statement

For a positive integer N>1N>1 with unique factorization N=p1α1p2α2⋯pkαkN=p_1^{\alpha_1}p_2^{\alpha_2}\dotsb p_k^{\alpha_k}, we define Ω(N)=α1+α2+⋯+αk.\Omega(N)=\alpha_1+\alpha_2+\dotsb+\alpha_k. Let a1,a2,…,ana_1,a_2,\dotsc, a_n be positive integers and p(x)=(x+a1)(x+a2)⋯(x+an)p(x)=(x+a_1)(x+a_2)\dotsb (x+a_n) such that for all positive integers kk, Ω(P(k))\Omega(P(k)) is even. Show that nn is an even number.
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