Question

This is bugging me, I feel like there's gotta be some simple answer but I can't seem to get it.

Answer 1

Show that if \(n\) is odd, then: \(\gcd(n,n+2)=1\)

Answer 2

oh i got it

Answer 3

umm okay how about we just think about in multiple sense

Answer 4

if its the the multiple of some prime u know that there has to be that prime difference atleast

Answer 5

nd the only prime difference with 2 differences are 2, and this is odd so

Answer 6

does that make sense?

Answer 7

so for example if u have 15 and 17
now 3 ,5 divide 15
but u know 3 cannot divide 17 because there a difference of 2
and 5 cannot dice it because again a diff of 2

Answer 8

similarly all the primes that divide some odd number cannot divide another odd number just 2 away, as they shud all have remainders building up as all the odd primes 3 or more

Answer 9

just think about multiples of primes

Answer 10

theres no way another multiple of the same prime will exist just 2 away

Answer 11

for odd ofc

Answer 12

GCD(n,n+2)=2, if n is even?

Answer 13

yup

Answer 14

I think I get what you're saying, like since the gap is smaller than 3 (the first prime they could share in common) then they have to be relatively prime?

I guess this makes sense but I don't know if I'm 100% convinced for some reason haha.

Answer 15

\[\gcd(n,n+2)=\gcd(n,2)=1,2, |n| \\ \text{ but } 2 \text{ is not a factor of } n \\ \text{ so } \gcd(n,2) \neq 2 \\ \\ \text{ in order for } \gcd(n,2)=|n| \\ \text{ we would have to have } \\ \text{ that } |n| \text{ is a factor of } 2 \\ \text{ which means } |n|=1 \text{ or } |n|=2 \\ \text{ but we already said } n \neq 2 \\ \text{ so } |n|=1 \]

Answer 16

yah

Answer 17

like it has to be true

Answer 18

think about it

Answer 19

Ok here I got a way of showing it that convinces myself:

Suppose \(p\) divides \(n\) and \(n+2\) then that means I can write \(n=pa\) and \(n+2=pb\) so:

\[n+2=n+2\]
\[pa+2=pb\]
that means \(p=2\) is the only prime that works, cause 2 has to also be divisible by p.

Answer 20

all other prime multiple of the other number will generate a remainder

Answer 21

okay yep thats pretty much same thing

Answer 22

so now u know there has to be this remainder

Answer 23

as p is greather than 2

Answer 24

Oh wow ok I should be thinking like @freckles is, I didn't even think about this:

\[\gcd(n,n+2)=\gcd(n,2)\]

ok in terms of remainders and the Euclidean algorithm this makes sense

Answer 25

oh i can see why that formula is true hehe :D

Answer 26

thanks everyone :D

Answer 27

pls medul

Answer 28

+fAN

Answer 29

hmmm... to medal dan or to not

Answer 30

that is the question

Answer 31

w o w

Answer 32

:D

Answer 33

learn to medal all nub

Answer 34

:D

Answer 35

pls im not an os hacker

Answer 36

dan likes medal parties

Answer 37

the more you reply here the higher your engagement score gets, probably, right?

Answer 38

I should suspend dan probably right

Answer 39

we can suspend him now and then unsuspend him in 30 cat seconds

Answer 40

So this has been solved?