## 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

**Download:** **PDF**

### Submission history

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

**Unique-IP document downloads:** 18 times

Vixra.org is a pre-print repository rather than a journal. Articles hosted may not yet have been verified by peer-review and should be treated as preliminary.
In particular, anything that appears to include financial or legal advice or proposed medical treatments should be treated with due caution.
Vixra.org will not be responsible for any consequences of actions that result from any form of use of any documents on this website.

**Add your own feedback and questions here:**

*You are equally welcome to be positive or negative about any paper but please be polite. If you are being critical you must mention at least one specific error, otherwise your comment will be deleted as unhelpful.*

*
*