Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

**juukii****Guest**

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

**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,626

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

Offline

**juukii****Guest**

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

**juukii****Guest**

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

**juukii****Guest**

No one can't help me?? ;(

**juukii****Guest**

Please help!!