Question

if gcd(a,b)=1, prove that gcd(a+b,b^2)=1

Answer 1

without saying i know what to do, i will make a guess
you know if \(p|ab\) then \(p|a\) or \(p|b\) since \(gcd(a,b)=1\)

then imagine \(p|a+b\) and also \(p|b^2\) and you want to show \(p=1\)
consider \((a+b)^2-b^2\) which \(p\) must divide as well

Answer 2

can you please give me a clearer hint?

Answer 3

well actually now that i look,that doesn't quite do it, does it?

Answer 4

i bet @joemath314159 has a better idea

Answer 5

earlier we proved that gcd(a,b^2) was one by showing there existed a linear combination of a and b^2 that equaled 1.

Answer 6

i also proved, using the same technique, gcd(a^2,b^2)=1

Answer 7

i was thinking along the lines of the proof that \(gcd(a+b,a^2+b^2)=1\)

Answer 8

actually 1 or 2

Answer 9

the hint my teacher gave me was to prove gcd(a+b,b)=1 first

Answer 10

ah, ok, i know the trick then, Since gcd(a^2,b^2)1, there exist integers x,y such that:\[a^2x+b^2y=1\Longrightarrow (a^2-b^2+b^2)x+b^2y=1\]\[\Longrightarrow (a^2-b^2)x+b^2(x+y)=1\Longrightarrow (a+b)\left[(a-b)x\right]+b^2(x+y)=1\]

Answer 11

There is your linear combination of a+b and b^2 that comes out to 1.

Answer 12

This is probably the dumbest way to do this problem though lol.

Answer 13

oh? is there another way to show this proof?

Answer 14

thanks a lot :D