Question

Can someone help me on the attached induction proof?

Answer 1

Not sure what to use as a base case

Answer 2

@ganeshie8 @mathmale @freckles

Answer 3

hey

Answer 4

for k is what you are messing with

Answer 5

k=0 is the base case

Answer 6

you want to show if m|(a-b) then we also have m|(a^0-a^0)

you want to assume we have m|(a-b) then we also have m|(a^n-b^n)
for some integer n>=0


then you want to use that assumption to show that m|(a-b) implies we also have
m|(a^(n+1)-b^(n+1))

Answer 7

@freckles but how does m|(a^0-a^0) = m|(a-b) ?

Answer 8

1-1=0
m|0

Answer 9

assume \[m|(a^n-b^n)\]
a^n-b^n=tm
where m is an integer.
\[a^n=tm+b^n\]
\[a ^{n+1}-b^{n+1}=a^na-b^nb=a(tm+b^n)-b^nb=atm+ab^n-b^nb=atm+(a-b)b^n\]
but m|(a-b) given
\[m|[atm+(a-b)b^n]\]
hence \[m|(a^{n+1}-b^{n+1})\]

Answer 10

correction t is an integer.

Answer 11

I don't understand the long equation a^n+1−b^n+1=a^na−b^nb ...

Answer 12

\[x^5=x^4 x\]