MathDB
Divisibility of a sum by a prime p

Source: Tuymaada 2012, Problem 4, Day 1, Juniors

July 21, 2012
quadraticsmodular arithmeticnumber theorynumber theory proposed

Problem Statement

Let p=1601p=1601. Prove that if 102+1+112+1+⋯+1(p−1)2+1=mn,\dfrac {1} {0^2+1}+\dfrac{1}{1^2+1}+\cdots+\dfrac{1}{(p-1)^2+1}=\dfrac{m} {n}, where we only sum over terms with denominators not divisible by pp (and the fraction mn\dfrac {m} {n} is in reduced terms) then p∣2m+np \mid 2m+n.
Proposed by A. Golovanov
Back to ProblemsView on AoPS