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2005 divides smth

Source: Moldova TST 2005

April 9, 2005
number theory unsolvednumber theory

Problem Statement

Let mNm\in N and E(x,y,m)=(72x)m+(72y)mxmymE(x,y,m)=(\frac{72}x)^m+(\frac{72}y)^m-x^m-y^m, where xx and yy are positive divisors of 72. a) Prove that there exist infinitely many natural numbers mm so, that 2005 divides E(3,12,m)E(3,12,m) and E(9,6,m)E(9,6,m). b) Find the smallest positive integer number m0m_0 so, that 2005 divides E(3,12,m0)E(3,12,m_0) and E(9,6,m0)E(9,6,m_0).