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a_n^2 = a_{n-1}a_{n+1} if a_2^2 = a_1a_3 where a_n=1+x^{n+1}+x^{n+2}

Source: Caucasus 2015 10.2

April 26, 2019
algebraidentityalgebraic identities

Problem Statement

Vasya chose a certain number xx and calculated the following: a1=1+x2+x3,a2=1+x3+x4,a3=1+x4+x5,...,an=1+xn+1+xn+2,...a_1=1+x^2+x^3, a_2=1+x^3+x^4, a_3=1+x^4+x^5, ..., a_n=1+x^{n+1}+x^{n+2} ,... It turned out that a22=a1a3a_2^2 = a_1a_3. Prove that for all n≥3n\ge 3, the equality an2=an−1an+1a_n^2 = a_{n-1}a_{n+1} holds.
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