Question

Prove for every integer (n), n^5-n = 0 (mod 30).

Answer 1

Have you considered factoring?

Answer 2

Factor it.

Answer 3

Why? Factor it.

Answer 4

@zepdrix

Answer 5

ohk ?

Answer 6

hihihi, I tried induction, but get stuck

Answer 7

n^5-n = 0 (mod 30).
u can use induction
also another way if u wanna use mod
n(n^4-1)=0 (mod 30)

chose one to start :D

Answer 8

for hypothesis step, I got (k-1)k(k+1)(k^2+1)
but the last step is k(k+1)(k+2)((k+1)^2+1)
then stuck, hihi

Answer 9

either way. Greedy me, I want to know both. hihi

Answer 10

mmm well lets induction it
for n=1 it work
lets assume for n
n^5-n = 0 (mod 30).
n( n^4-1) = 0 (mod 30).
n( n+1)(n-1)(n^2+1) = 0 (mod 30).
is true
prove for n+1
(n+1)((n+1)^2-1) ((n+1)^2+1)

lol im not getting the induction step xD

Answer 11

neither I, That's why I tagged others to solve it. haiz. it's not a piece of cake.

Answer 12

xD

Answer 13

lets dump the induction and try some number theory

Answer 14

n( n^4-1) = 0 (mod 30).
if this one is true then ..
n( n^4-1) = 0 (mod 2*3*5).

Answer 15

It's up to you . I don't know number theory.:)

Answer 16

ok did u got that step ?

Answer 17

among the hypothesis, we have (n-1)(n)(n+1) , they are 3 consecutive integer, so that they divided by 2 for sure :)

Answer 18

so we have 3 equtions
n( n^4-1) = 0 (mod 2 ).
n( n^4-1) = 0 (mod 3).
n( n^4-1) = 0 (mod 5).

Answer 19

good u already proved it for 2 :D

Answer 20

nw for 3
any number would devide 3 should be of the form 3k right ?

Answer 21

using devisition algorithm n have 3 choices
n=3m
n=3m+1
n=3m+2

Answer 22

got it tell nw ?

Answer 23

let finish the proof. I will ask if I have question

Answer 24

got it

Answer 25

so for n=3m
n( n^4-1) = 0 (mod 3 )
3m(....)=0(mod 3 )
devide 3

for n=3m+1

(3m+1)(3m+2)(3m) ((3m)^2+1)) = 0 (mod 3 )

as u can see there is 3m so devide 3

Answer 26

for (3m+2)
n( n+1)(n-1)(n^2+1)
(3m+2)(3m+3) (....)=0mod 30
3(...) =0mod 30
so it devide 3

Answer 27

I got this part (prove the expression divided by 3) how about 5

Answer 28

ok about 5 mmm
the numer devide 5 when first digit =0,5
but that wont help in proof

Answer 29

other case the number devide 5 when its of the form 5G
we should use D.A again :D for g=5
so n would have 5 cases
n=5k
n=5k+1
n=5k+2
n=5k+3
n=5k+4

Answer 30

ok, I got 5 too. Thank you hihihi

Answer 31

However, my question is about how to combine them?

Answer 32

lolz whats hihihih :P

Answer 33

combine what ?

Answer 34

since the proof of 2 include in 3, (3 consecutive numbers) but for 5, we assume that n = 5k,..... which is not include 2 and 3 in

Answer 35

well its a theorm that set of \( d_0 ,d_1, d_2 ...\) are devisors of a number g
and
n = 0 mod g
then
n= 0 mod \( [d_0×d_1× d_2×...]\)

Answer 36

i dint got ur question , do u wanna prove it for 5 ?

Answer 37

Yes, but n is vivid
you set n =5m , and we have 5m divided by 5, but how it divided by 3 also?

Answer 38

O.O

Answer 39

u said u got that part

Answer 40

I am sorry for my bad English to interpret my mind. let me arrange it
for 3, you set n =3m, n = 3m+1, n=3m+2
and our expression: (n-1)(n)(n+1) (n^2+1) , now replace n = 3m for example,
we still have 3m (3m -1)(3m+1) (this part is 3 consecutive numbers also, Therefore, this part divided by 2 also.That's what I mean divided by 2 is include the part of divided by 3)
but divided by 5 is another problem. It included 2 part, but not 3 .

Answer 41

mmm lol
ok u dint get the concept of D.A when i set this
n^5-n = 0 (mod 3 ).
then forget about 2 or 5
im only trying to prove for 3

Answer 42

and since u dnt know this concept i suggest not to continue ... and try something else

Answer 43

what is your course name ?

Answer 44

can you give me a link to learn more ?

Answer 45

i might find some other method

Answer 46

yeah sure

Answer 47

wait

Answer 48

hihihi.... it's not my problem. it's my curiosity.

Answer 49

wow !
where did u find it ?
how old are you ?

Answer 50

this post is not mine, either. hihihj

Answer 51

ohh ic !
well ull gonna find it easy !

Answer 52

here , go through chapter 1 and 2
https://drive.google.com/file/d/0B1oczGnMVhNqTDMtTEs0QzFPTm8/edit?usp=sharing

Answer 53

let me know if u got it , or if u need something else :D

Answer 54

thank you

Answer 55

np :)

Answer 56

\(n\cdot(n-1)\cdot(n+1)\cdot (n^{2}+1)\)

Obviously, with three consecutive integers, n-1, n, and n+1, this thing is divisible by 2 and by 3. The only challenge is 5.

Can we prove that either \(n^{3} + n\) or \(n^{3} - n\) is divisible by 5?

Answer 57

@tkhunny
there are two ways to prove it in number theory
1_ using difision algorithm for 5and try 5 possible values
n=5m
n=5n+1
n=5n+2
n=5m+3
n=5m+4
n=5m+5
2_there is a theorim Euler that sit
for any number a and prime P , gcd(a,p)=1
then
a^5=a mod 5
so
a^5-a=0 mod 5
done