Question

Show that if \(1 = as + bt\), \(\gcd(a,b) = 1\)

Answer 1

So actually is it just the Bezout's Lemma?

Answer 2

\[\gcd(a,b) = as + bt \]

Answer 3

\[1 = \gcd(a,b) = as + bt\]So,\[1 = \gcd(a,b)\]That's the only work?

Answer 4

greatest common divisor?

Answer 5

yes

Answer 6

its a theorem of number theory right?

Answer 7

Yes, elementary number theory :)

Answer 8

ya..........its a result that when and b are relatively prime .....then there exists integers s and t such that 1=as +bt

Answer 9

gcd(a,b) = 1, then gcd(a^2,b^2)=1.

Since both = 1 we can write this as
gcd(a,b) = gcd(a^2,b^2)

Then a = a^2
and b = b ^2

There are only 2 real numbers that when squared equals itself, 0 and 1.

Since we are assuming that both equations in fact = 1, then a and b both equal one, If either gcd (a,b) or gcd(a^2,b^2)= 0 the only other choice, then gcd (a,b) or gcd(a^2,b^2) would not equal 1.

Answer 10

What? o.O

Answer 11

I asked if my statement was enough. lol

Answer 12

special case of Bezuot's identity.

Answer 13

Exactly -- can you see my solution?

Answer 14

looks like both works either way!! that should be right!!

Answer 15

Okay :)

Answer 16

although proving this lemma is fairly complicated.