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 1 of 10: Extract sum and difference divisibilities

Previous step Next step

c=100!,\qquad a_{r+s}\mid c(a_r+a_s),\qquad a_{r-s}\mid c(a_r-a_s)\ (r>s)

Detailed analysis

Put $c=100!$ . The original divisibility condition first yields the two linear divisibilities displayed in this step. If a common divisor divides every term, divide all terms by it; these two derived divisibilities are preserved. Thus we may assume the normalized sequence has gcd $1$ and restore the common factor at the end. Comparing the conditions for $[r,s]$ and $[r+1,s-1]$ when $s\ge r+2$ gives $a_{r+s}\mid c(a_r+a_s)$ ; the cases $s=r$ and $s=r+1$ give the same conclusion directly. Comparing $[s,r-1]$ and $[s+1,r]$ gives $a_{r-s}\mid c(a_r-a_s)$ for $r>s$ .