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Loser66
 one year ago
How to prove a^3 a divisible by 3 for all a?
Please, help
Loser66
 one year ago
How to prove a^3 a divisible by 3 for all a? Please, help

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SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1Factors into: *a(a1)(a+1)*

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0I know a^3 a = a(a+1)(a1) are 3 consecutive numbers, hence it divisible by3

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1so what's the prob/

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0All I want is to go this way a is odd, a = 2k +1 a is even, a = 2k

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0but I got stuck at a is odd.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0It is back to 3 consecutive numbers.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1whether you start from odd or even integer, the product of 3 consequtive integers will be divisible by 3....

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0:) Hence, no way to avoid "3 consecutive numbers" method?? right?OOOOOOk, that's all I want to know. Thank you.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1Yes, 3 conseq.... (or not another way that I aware of using my small knowledge)

freckles
 one year ago
Best ResponseYou've already chosen the best response.0you could do induction if you don't like that method above... \[\text{ assume } 3i=a^3a \text{ for some integer } a \\ ... \\ (a+1)^3(a+1)=a^3+3a^2+3a+1a1 \\ (a+1)^3(a+1)=a^3a+3a^2+3a \\ (a+1)^3(a+1)=3i+3a^2+3a \\ (a+1)^3(a+1)=3(i+a^2+a) \\ \text{ so } 3 \text{ is a factor of } (a+1)^3(a+1)\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.0and i is an integer of course

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0Got you. Thanks @freckles
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