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## #1 2005-11-02 08:03:11

juukii
Guest

### Hard

We have  p > 3  where "p" is number first and a,b,c which are all-out and positive and we know that    a + b + c = p + 1
and a³ + b³ + c³ - 1  is divisible by "p".
Prove that one of a,b or c     equal 1

## #2 2005-11-02 08:40:51

MathsIsFun
Registered: 2005-01-21
Posts: 7,664

### Re: Hard

I have an idea for a start.

Using: a³ + b³ + c³ - 1  is divisible by "p"
Perhaps only a³ + b³ can be divisible by p, so subtracting 1 requires adding 1, and that is the c³ term

Also:
a + b + c - 1 = p
( a³ + b³ + c³ - 1 )  /  ( a + b + c - 1 ) = whole number

That is as far as I have got, and I have to go do something else now.

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

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## #3 2005-11-02 09:38:51

juukii
Guest

### Re: Hard

I've tried to solve it in many ways but I still have nothing...

## #4 2005-11-04 02:48:09

juukii
Guest

### Re: Hard

Is there anybody who can solve it...please it's important

## #5 2005-11-04 23:40:54

juukii
Guest

### Re: Hard

No one can't help me?? ;(

juukii
Guest