Number Theory


2 to the Power of (p-1) is not Congruent to 1 Mod P Cubed for Any Prime 'p'

Authors: Ramaswamy Krishnan

If\quad { 2 }^{ p-1 }\quad \equiv \quad 1\quad mod\quad ({ p }^{ 3 })\quad then\quad 2,-1,{ 2 }^{ p-2 }\quad are\quad solutions\quad to\quad the\quad equation\\ f(a)\quad =\quad 1\quad -\quad { a }^{ p }\quad -\quad { (1-a) }^{ p\quad \quad }\equiv \quad o\quad mod({ p }^{ 3 }).\quad Using\quad this\quad fact\quad and\quad an\quad expression\quad for\\ { (x+y) }^{ n }\quad \quad in\quad terms\quad of\quad xy\quad ,\quad (x+y)\quad ,\quad ({ x }^{ 2 }+xy+{ y }^{ 2 })\quad it\quad is\quad prooved\quad that\\ { 2 }^{ p-1 }\quad \ncong \quad 1\quad mod({ p }^{ 3 })\quad for\quad any\quad prime\quad 'p'.

Comments: 3 Pages. Title has been changed a little bit. Instead of mod p, it should be mod p cubed

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Submission history

[v1] 2017-08-17 05:19:11

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