Problem 5

Consider an infinite sequence

a_1,a_2,\ldots

of positive integers such that

100!(a_m+a_{m+1}+\cdots+a_n)

is a multiple of

a_{n-m+1}a_{n+m}

for all positive integers

m,n

with

m\le n

. Prove that the sequence is either bounded or linear. A sequence is bounded if there is a constant

N

such that

a_n<N

for every positive integer

n

, and it is linear if

a_n=n\cdot a_1

for every positive integer

n

Step 10 of 10: The persistent prime contradicts the normalization

Previous step Next step

a_n\mid c(a_{n+k}-a_k),\quad a_{n+k}\mid c(a_n+a_k)\Longrightarrow p\mid a_k\quad\forall k

Detailed analysis

Fix any $k$ . Since $a_n\mid c(a_{n+k}-a_k)$ by Step 1, unboundedness lets us choose $n$ so large that $a_n$ and $a_{n+k}$ are at least $a_r$ (for sufficiently large choices, positivity forces the difference to be positive; indeed $a_{n+k}\ge a_n/c-a_k$ ). Hence both are divisible by $p^{e'}$ . The other relation in Step 1, $a_{n+k}\mid c(a_n+a_k)$ , and $e'>v_p(c)$ now imply $p\mid a_k$ . Since $k$ was arbitrary, $p$ divides every term, contradicting the normalization $\gcd(a_1,a_2,\ldots)=1$ . Therefore the sequence is bounded in this case.