Zhu Wen-yu
Abstract
The author proves the congruence 2n≡5(mod n)(n1) except the trivial solution n=3,only one solution n=19147=41·467 in the interval [2,4294967295].And when m1 satisfy the congruence 2m≡5(mod m) then n=2m-1 is a solution of the congruence 2n-4≡1(mod n).
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The author proves the congruence 2n≡5(mod n)(n1) except the trivial solution n=3,only one solution n=19147=41·467 in the interval [2,4294967295].And when m1 satisfy the congruence 2m≡5(mod m) then n=2m-1 is a solution of the congruence 2n-4≡1(mod n).
Key concepts: Mod, Congruence (geometry), Mathematics, Congruence relation, Combinatorics, Geometry