Question 29415: show that 3 is a factor of n^3+2n for all positive integers n.
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! show that 3 is a factor of n^3+2n for all positive integers n
HOPE YOU WERE TAUGHT FERMAT'S THEOREM....IT SAYS IF P IS A PRIME NUMBER THEN
N^P=N(MOD P)...SINCE 3 IS A PRIME NUMBER WE HAVE
N^3=N(MOD 3)
SO 3 DIVIDES N^3-N
SO 3 DIVIDES N^3-N+3N SINCE 3N IS DIVISIBLE BY 3
SO 3 DIVIDES N^3+2N.....
IF YOU ARE NOT TAUGHT THIS THEOREM ..COME BACK...I SHALL GIVE ALTERNATIVE...
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ok ,let us prove it without using fermats theorem
SINCE WE HAVE THE NUMBER AS 3 WHICH IS VERY SMALL ,WE NEED NOT GO FOR
COMPLICATED METHODS ,BUT CAN USE A SIMPLER METHOD.INDUCTION CAN ALSO
BE USED.
LET US SEE
METHOD 1.....
T.S.T .....N^3+2N IS DIVISIBLE BY 3....
LET US PUT ...........Z=N^3+2N
WHEN A NUMBER IS DIVIDED BY 3 WE CAN HAVE 3 POSSIBLITIES ON REMAINDER...
1.REMAINDER IS ZERO ...THAT IS THE NUMBER IS A MULTIPLE OF 3 ...N = 3
X WHERE X IS AN INTEGER..SO
Z=N^3 +2N = (3X)^3+2*3X= 27X^3+6X=3(9X^2+2X) = 3*M
WHERE M = 9X^2+2X IS AN INTEGER...SINCE X IS AN INTEGER .
HENCE Z IS DIVISIBLE BY 3.
2.REMAINDER IS ...1...THAT IS THE NUMBER N =3X+1 WHERE X IS AN INTEGER.
Z=(3X+1)^3+2(3X+1) = 27X^3+27X^2+9X+1+6X+2
=3(9X^3+9X^2+5X+1)=3*M WHERE
M=9X^3+9X^2+5X+1 IS AN INTEGER SINCE X IS AN INTEGER.
HENCE Z IS DIVISIBLE BY 3.
3.REMAINDER IS ...2..THAT IS THE NUMBER N =3X+2 WHERE X IS AN INTEGER.
Z=(3X+2)^3+2(3X+2) = 27X^3+54X^2+36X+8+6X+4
=3(9X^3+18X^2+12X+4)=3*M WHERE
M=9X^3+18X^2+12X+4 IS AN INTEGER SINCE X IS AN INTEGER.
HENCE Z IS DIVISIBLE BY 3.
HENCE Z IS ALWAYS DIVISIBLE BY 3 FOR ANY INTEGRAL VALUE OF N.
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INDUCTION METHOD OF PROOF....
IN INDUCTION WE FOLLOW 4 STEPS...
I STEP...WE CHECK WHETHER THE THEOREM IS TO BE PROVED TRUE FOR ALL
NATURAL NUMBERS N...
YES ...SO...OK...
II STEP..WE CHECK WHETHER THE GIVEN STATEMENT IS TRUE FOR N=1
ZN=N^3+2N IS DIVISIBLE BY 3 IS THE THEOREM..SO FOR
N = 1 WE HAVE
Z1=1^3+2*1=3...IS DIVISIBLE BY 3 ...OK.
III STEP...WE ASSUME IT IS TRUE FOR N=K AN INTEGER...THAT IS...
ZK = K^3+2K.. IS DIVISIBLE BY 3...ASSUMED..
THAT IS ...ZK = K^3+2K=3M... WHERE M IS AN INTEGER.
IV STEP...IF III IS TRUE THEN ,WE TRY TO PROVE THAT IT IS ALSO TRUE
FOR N=K+1...THAT IS TO PROVE THAT
Z(K+1)=(K+1)^3+2(K+1) IS DIVISIBLE BY 3..
Z(K+1)=(K+1)^3+2(K+1)=K^3+3K^2+3K+1+2K+2
=(K^3+2K)+3(K^2+K+1)=3M+3*INTEGER ....SINCE K^2+K+1 IS AN INTEGER,K
BEING AN INTEGER...HENCE
Z(K+1)=3*(M+INTEGER )= 3 * INTEGER
THAT IS Z(K+1) IS DIVISIBLE BY 3 IF ZK IS DIVISIBLE BY3
CONCLUDING ARGUMENT....
WE PROVED IN II STEP THAT THE THEOREM IS TRUE FOR K=1..BY IV STEP,THEN
IT IS TRUE FOR K+1 =1+1 =2...THEN IT IS TRUE FOR K+1=2+1=3......SO ON
...HENCE IT IS TRUE FOR ALL NATURAL NUMBERS N...
> ok ,let us prove it without using fermats theorem
> SINCE WE HAVE THE NUMBER AS 3 WHICH IS VERY SMALL ,WE NEED NOT GO FOR
> COMPLICATED METHODS ,BUT CAN USE A SIMPLER METHOD.INDUCTION CAN ALSO
> BE USED.
> LET US SEE
> METHOD 1.....
> T.S.T .....N^3+2N IS DIVISIBLE BY 3....
> LET US PUT ...........Z=N^3+2N
> WHEN A NUMBER IS DIVIDED BY 3 WE CAN HAVE 3 POSSIBLITIES ON REMAINDER...
> 1.REMAINDER IS ZERO ...THAT IS THE NUMBER IS A MULTIPLE OF 3 ...N = 3
> X WHERE X IS AN INTEGER..SO
> Z=N^3 +2N = (3X)^3+2*3X= 27X^3+6X=3(9X^2+2X) = 3*M
> WHERE M = 9X^2+2X IS AN INTEGER...SINCE X IS AN INTEGER .
> HENCE Z IS DIVISIBLE BY 3.
> 2.REMAINDER IS ...1...THAT IS THE NUMBER N =3X+1 WHERE X IS AN INTEGER.
> Z=(3X+1)^3+2(3X+1) = 27X^3+27X^2+9X+1+6X+2
> =3(9X^3+9X^2+5X+1)=3*M WHERE
> M=9X^3+9X^2+5X+1 IS AN INTEGER SINCE X IS AN INTEGER.
> HENCE Z IS DIVISIBLE BY 3.
>
> 3.REMAINDER IS ...2..THAT IS THE NUMBER N =3X+2 WHERE X IS AN INTEGER.
> Z=(3X+2)^3+2(3X+2) = 27X^3+54X^2+36X+8+6X+4
> =3(9X^3+18X^2+12X+4)=3*M WHERE
> M=9X^3+18X^2+12X+4 IS AN INTEGER SINCE X IS AN INTEGER.
> HENCE Z IS DIVISIBLE BY 3.
> HENCE Z IS ALWAYS DIVISIBLE BY 3 FOR ANY INTEGRAL VALUE OF N.
> I SHALL GIVE YOU SEPERATE MAIL ON INDUCTION..
> VENUGOPAL.
>
>
>
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