MathLabs

Problem 3

For each integer k≥2k\ge2, determine all infinite sequences of positive integers a1,a2,…a_1,a_2,\ldots for which there exists a polynomial PP of the form P(x)=xk+ck−1xk−1+⋯+c1x+c0P(x)=x^k+c_{k-1}x^{k-1}+\cdots+c_1x+c_0, where c0,c1,…,ck−1c_0,c_1,\ldots,c_{k-1} are non-negative integers, such that P(an)=an+1an+2⋯an+kP(a_n)=a_{n+1}a_{n+2}\cdots a_{n+k} for every integer n≥1n\ge1.
Step 2 of 6: A telescoping identity
In plain words

Writing the defining relation for two consecutive indices and dividing cancels most of the product, leaving a clean relationship between ana_n and an+ka_{n+k}.

P(an)P(an−1)=an+kan\dfrac{P(a_n)}{P(a_{n-1})}=\dfrac{a_{n+k}}{a_n}
Detailed analysis

The relation gives P(an−1)=anan+1⋯an+k−1P(a_{n-1})=a_na_{n+1}\cdots a_{n+k-1} and P(an)=an+1an+2⋯an+kP(a_n)=a_{n+1}a_{n+2}\cdots a_{n+k}. Dividing, all the shared factors an+1,…,an+k−1a_{n+1},\ldots,a_{n+k-1} cancel, leaving P(an)/P(an−1)=an+k/anP(a_n)/P(a_{n-1})=a_{n+k}/a_n.