Question

show the n^5 - n is divisible by 3 for all integers n

Answer 1

please help! :)

Answer 2

Let's try to factor n^5 - n.
I get (n - 1)n(n + 1)(n^2 + 1)

Answer 3

One in three consecutive integers MUST be divisible by 3.

Answer 4

im sorry i dont get the first equation but i get ur 2nd comment

Answer 5

Do you know factoring? Just try a calculator or something

Answer 6

so basically you've factorised it again and again?

Answer 7

so the first factorisation is n(n^4 - 1)?

Answer 8

\(\Rightarrow n(n^4 - 1) \)

\(\Rightarrow n( (n - 1) (n^3 + 3n^2 + 3^2n + 1)) \)

Answer 9

yep :)
i got that

Answer 10

But I want to cut down what I am explaining you. Leave the factorization for a while.

Answer 11

ok

Answer 12

Yay! Now, if you notice, one in three consecutive integers can be divided by 3.

Answer 13

yep

Answer 14

See how there's this thing (n - 1)n(n + 1)?? So that means we are multiplying three consecutive integers.

Answer 15

That means that it is divisible by 3, as we can divide one of them and we get a proper quotient :)

Answer 16

wait sorry just a minute :)

Answer 17

where did u get (n - 1)n(n + 1) from again? it is factorisation?

Answer 18

Yeah ^^^ see there... the factorization

Answer 19

The factored form is (n - 1)n(n + 1)(n^2 + 1), but if we focus on (n - 1)n(n + 1) we see three consecutive integers.. and see the rest that I explained.

Answer 20

ohh i get it so if say n is 3 than its 2, 3, and 4?

Answer 21

thanks :) i get it now..but what do i do now that i have that?

Answer 22

Yeah, so 3 is divisible by 3 right? :)

Answer 23

in the answer i just say the because of this factorisation it is divisible by 3

Answer 24

(2)(3)(4)(10) = 240 a proof.

Answer 25

yep i get that 3 is divisible by 3

Answer 26

No.. it is because (n - 1)(n)(n + 1) signifies the multiplication of 3 consecutive integers, and one in three consecutive integers is divisible by 3 =)

Answer 27

ahhhh so the actually factorisation is ^^^ and when u substitute n with a number say 3 then u get a number that is divisible by 3 right?

Answer 28

actual*

Answer 29

No, proofs don't involve trying it with numbers, as proofs may have exceptions.

Answer 30

ohh okay

Answer 31

so for the answer i can put the equation n^5 - n

then i put the factorised bits (n - 1)n(n + 1)(n^2 + 1)

and then i say that because they are consecutive they are divisible by 3?

Answer 32

but just one thing..if there are 3 consecutive integers (n - 1)(n)(n + 1) how is it one in three consecutive integers is divisible by 3?

Answer 33

would this deal with .... forgot the name of it ...

Answer 34

if its true for k, then its assume its true for k+1 etc

Answer 35

induction right? yea i dont get that... lol :)

Answer 36

induction :) thats what i was thinking

Answer 37

can u explain induction to me?

Answer 38

or is what i was doing before right>

Answer 39

n^5 - n

there is nothing obvious that would make this go bad for integers; so lets set up a few basis cases

dunno if the prior method was viable or not

Answer 40

ok..so what do i do? im sorry im soo stuck... :P

Answer 41

0^5 - 0 = 0 ; 0/3 is good

1^5 - 1 = 0 ; 0/3 is good

2^5 - 2 = 30; 30/3 is good

assume its good for k

k^5 - k is divisible by 3 is the assumption from the basis

Answer 42

hmm, what to do next

Answer 43

oh ok i get that..k is just a random letter replacing n

Answer 44

correct

Answer 45

next we would change k to k+1 and work it out

Answer 46

okay so it wud be (k + 1)^5 - (k+1)?

Answer 47

yes

Answer 48

now to deduce of thats divisible by 3 :)

Answer 49

ok so how do i do that :P

Answer 50

cant recall excatly, but id prolly start by expanding the whole thing

Answer 51

Woah woah.. we have already proved the existence of the formula in n ;)

Answer 52

1^5 - 1 = 1 - 1 = 0 that is divisible by 3.

Answer 53

And then we just deduced it for n.. and we can just in the same way deduce for n + 1

Answer 54

:P yea u have lol but i dont get how its to do with one in three consecutive integers is divisible by 3

Answer 55

Where don't you understand?

Answer 56

k^5 + 5k^4 + 10k^3 + 10k^2 +5k + 1 - k - 1

k^5 + 5k^4 + 10k^3 + 10k^2 +5k - k

k^5 + 5k^4 + 10k^3 + 10k^2 + 4k

hmmm

Answer 57

If n is divisible by 3, then the quotient is (n + 1)(n - 1)(n^2 + 1)
If n + 1 is divisible by 3, then the quotient is n(n - 1)(n^2 + 1)..
If n - 1 is divisible by 3, then the quotient is n(n + 1)(n^2 + 1)
You can even try out the Polynomial Remainder theorem, and you'd see that it has no remainder in all three cases.

Answer 58

ohh okay thanks :)

Answer 59

i believe parth has it, but i cant recall too well if thats adequate for the induction or not :/

Answer 60

Both work

Answer 61

I GOT IT NOW!! u guys are awesome! :D

Answer 62

im kinda slow :P

Answer 63

:) good luck

Answer 64

lol

Answer 65

i got it by substituting the n in n(n - 1)(n + 1)(n^2 + 1) by 2 and then getting 30 which id divisible by 3

Answer 66

if 2 numbers are divisible by 3; then the sum is also divisible by 3 ....

Answer 67

that is what i was thinking :)

Answer 68

so i prove it with 2 numbers than it works with all numbers yay!

Answer 69

iv been stuck on that question for 3 hours :P

Answer 70

prove it with a basis case; then assume it works for all cases of "k"; then add (k+1) to both sides and algebra it into submission is the usual induction steps that I recall

Answer 71

i was stuck changing a flat tire for like 3 hours :)

Answer 72

hahah welll its 11:30 and i have school tomorrow so i better go to sleep :) thanks for ur help!

Answer 73

yw, and again ... good luck

Answer 74

thanks :)