Question

Show that x+y is a factor of x^(2n-1)+y^(2n-1) for all natural numbers n.

Use mathematical induction.

Answer 1

Why do we have to do these by induction? =/ we could just use the binomial theorem and be done in a line or two.

Answer 2

My professor requires all of them to be solved with induction :\

Answer 3

I'll assume you can do the Base Step in the inductive process, so i'm going to start with the Inductive step (assuming the statement is true for n = k, and using this to prove the statement is true for n = k+1)

Answer 4

is that ok? or do we need to go over the Base Step too?

Answer 5

That's fine. I have the base step done.

Answer 6

if n = 0 does it hold?

Answer 7

ok, So lets assume for some fixed natural number k that:
\[(x+y)\mid x^{2n-1}-y^{2n-1}\] We want to show that:
\[(x+y)\mid x^{2n+1}-y^{2n+1}\]

Answer 8

That notation\[a\mid b\]means "a divides b" btw, i dont know if thats clear or not.

Answer 9

when n = 0
\[\frac{1}{x}+\frac{1}{y} = \frac{x + y}{xy}\]

Answer 10

Using our Inductive Hypothesis, we have:\[(x+y)\mid x^{2k-1}-y^{2k-1}\iff x^{2k-1}-y^{2k-1}=b(x+y)\]for some integer b. Solving for x^(2k-1) gives:
\[x^{2k-1}=b(x+y)+y^{2k-1}\]
Now looking at the statement:\[x^{2k+1}-y^{2k+1}\]we can rewrite it as:\[x^2\cdot x^{2k-1}-y^{2k+1}\]

Answer 11

Now, we are going to take our Inductive Hypothesis and substitute it into that statement.

Answer 12

So that nets us:
\[x^2(b(x+y)+y^{2k-1})+y^{2k+1}\]
Now, the name of the game is to see if we can manipulate this statement to become something like:
\[(x+y)(junk)\] the junk part is going to look ugly, but if we can pull out a (x+y) like that we are good.

Answer 13

\[\frac{b(x+y)*x^2y+x^2y^{2k}+y^{2k+2}}{y}\]

Answer 14

i dont see it yet >.< do you Satellite?

Answer 15

How can we get that x+y out of there

Answer 16

oh i wasn't looking. just trying to recall how to prove all horses are the same color by induction. i know how to prove you will not be hanged by induction

Answer 17

lol

Answer 18

jailer says you will be hung before saturday but you will not know the day in advance. prisoner says ok well if i will not know the day in advance then they cannot wait until saturday because that is the last day, and i will know on friday that they will hang me saturday. but if they cannot hang me on saturday surely they cannot hang me on friday, because since i will know the day before, and i can't be hung on saturday, then i cannot be hung on friday. etc.

Answer 19

oh im so stupid >.< its because its late. I made a sign error typing it out. it should be:
\[x^{2k+1}-y^{2k+1}=x^2(x^{2k-1})-y^{2k+1}\]\[=x^2(b(x+y)+y^{2k-1})-y^{2k+1}=bx^2(x+y)+x^2y^{2k-1}-y^{2k+1}\]\[bx^2(x+y)+y^{2k-1}(x^2-y^2)=bx^2(x+y)+y^{2k-1}(x+y)(x-y)\]

Answer 20

i think the gimmick is to factor out and subtract off. something like
\[x^{2n+1}+y^{2n+1}=(x^{2n-1}+y^{2n-1})x^2y^2-stuff\]

Answer 21

Now you can factor out the (x+y) and you get:
\[(x+y)(bx^2+y^{2k-1}(x-y))\]
This concludes my Proof by Mathematical Induction.
(stupid sign errors >.>)

Answer 22

Well done professor Joesph!!!!! and hi joe

Answer 23

yo :)

Answer 24

Satellite! i need you help with a homework problem lol ima post a new question.

Answer 25

im gonna scan it, one sec.

Answer 26

ok but it may have to wait until the morning. let me see i can do this one

Answer 27

professor needs help from LEGAND COOL

Answer 28

\[x^{2n+1}-y^{2n+1}=(x^{2n-1}-y^{2n-1})x^2y^2-stuff\]

Answer 29

If all else fails... remember to use logarithms.

Answer 30

now what is stuff?

Answer 31

im gonna post a new question., posting....now <.<

Answer 32

ah i can't do this it is too late. i will look at yours in the morning

Answer 33

lol fine <.<

Answer 34

\[x^{2k+1}+y^{2k+1}=\]\[=x^{2k+1}+x^2y^{2k-1}-x^2y^{2k-1}-x^{2k-1}y^2+x^{2k-1}y^2+y^{2k+1}=\]
\[=x^2(x^{2k-1}+y^{2k-1})-x^2y^2(x^{2k-3}+y^{2k-3})+y^2(x^{2k-1}+y^{2k-1})=\]
\[=(x^2+y^2)(x^{2k-1}+y^{2k-1})-x^2y^2(x^{2k-3}+y^{2k-3})\equiv 0\,\,\,(mod\,\,(x+y))\]