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sum of any 41 consecutive numbers on this circle is a multiple of 41^2

Source: Tournament of Towns, Junior O-Level , Fall 2019 p4

April 19, 2020
number theoryconsecutivemultiplecircleSum of Squares

Problem Statement

There are given 10001000 integers a1,...,a1000a_1,... , a_{1000}. Their squares a12,...,a10002a^2_1, . . . , a^2_{1000} are written in a circle. It so happened that the sum of any 4141 consecutive numbers on this circle is a multiple of 41241^2. Is it necessarily true that every integer a1,...,a1000a_1,... , a_{1000} is a multiple of 4141?
(Boris Frenkin)
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