Question

How do I show the proof for: For all integers a, if a^2 = 0(mod 6), then a = 0(mod 6)? I think the answer is true, but I don't know how to prove it generally.

Answer 1

a=0(mod 6)
multiplying both sides by a give us :
a^2=0(mod 6)

Answer 2

thats the converse direction,
but we wanto prove this : given a^2=0 mod 6, prove a = 0 mod 6

Answer 3

let me try :) @ganeshie8 watch me, please :)
a^2 =0 (mod 6)
a^2-0 =0 (mod6)
(a+0)(a-0) =0 (mod 6)
a+0 =0 (mod6)
a =0 (mod 6)

Answer 4

:)

Answer 5

looks legit :) u have used quadratic congruences !

Answer 6

another long/boring way is :

a^2 = 0 mod 6
=>
a^2 = 0 mod 2
a^2 = 0 mod 3

2 | a^2 => 2 | a
3 | a^2 => 3 | a

since gcd(2,3) = 1,
2*3 | a

a = 0 mod 6

Answer 7

hihihi....

Answer 8

Like you did 2 | a^2 => 2 | a, why not 6 | a^2 => 6 | a?

Answer 9

6 is not prime

Answer 10

OK, I got it!

Answer 11

im looking for a contradiction to show n | a^2 does not necessarily mean n | a
but not getting any quick examples...

Answer 12

n is composite..

Answer 13

a = b (mod n)

division is not allowed here unless 'a' and 'n' are coprime... so it shouldnt work in general....

Answer 14

0 1 2 3 4 5
0 0 0 0 0 0 0
1 0 1 2 3 4 5
2 0 2 4 0 2 4
3 0 3 0 3 0 3
4 0 4 2 0 4 2
5 0 5 4 3 2 1

0^2 = 0
1^2 = 1
2^2 = 4
3^2 = 3
4^2 = 4
5^2 = 1

proof by table

Answer 15

:)

Answer 16

learn every day!!

Answer 17

@ganeshie8 you said division is not allowed, so how do you consider the Quadratic congruence as you call it accepted ?

Answer 18

direct division is not allowed,

division is defined as below :-
a = b mod n

a/c = b/c mod (n/gcd(c, n) )

Answer 19

when u divide, u need to divide the mod also by gcd(c, n)
oly dividing the remainders will not work

Answer 20

(a+0)(a-0) =0 (mod 6) to a+0 =0 (mod6) isn't a direct division ?

Answer 21

thats just zero product property which is legal

Answer 22

it is legal in congruences also, not just algebra..

Answer 23

ab = 0
=>
either one of them or both are 0

Answer 24

Ok

Answer 25

here is one contradicting example for the statement :
a^2 = 0 mod n does not necessarily mean a = 0 mod n

4^2 = 0 mod 16 , but 4 = 0 mod 16 is false.

Answer 26

isn't mod 6

Answer 27

we're talking in general

Answer 28

not specific to a 6 or something...
6 is not a special number... not worth remembering its properties by heart :)

Answer 29

good, How do you prove that 1^n+2^n+3^n+4^n=0 mod 5 ?

Answer 30

easy :
4 = -1
3 = -2

everything cancel out

Answer 31

u doing number theory or groups ?

Answer 32

or 1 and 2 congruent to -3 and -4 right ?

Answer 33

yes,
1 = -4
2 = -3

Answer 34

so it means always the number of sum parts must be even ?

Answer 35

im not sure we can generalize like that... wat do we get by generalizing that anyway ?

Answer 36

try this :
find the remainder when \(\large 86^{1234921897129}\) is divided by 87

Answer 37

yup ! too easy for u Loser :)

Answer 38

nobody can help u cuz u are doing all tough problems nobody ever messed wid lol