Suppose gcd(a,b) = 1, prove gcd(na,nb) = n
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Hello qtpipi Originally Posted by qtpipi Suppose gcd(a,b) = 1, prove gcd(na,nb) = n
Any help? is a common divisor of and . Suppose now that . Then, since every common divisor of and is a divisor of , for some and .
So is a common divisor of and . Contradiction, since .
Thank you thank you!
I have another similar problem:
Prove that gcd(a+b, a-b) = gcd(2a, a-b) = gcd(a+b, 2b)
Any help with this one too??? Thanks!
Nevermind I figured it out!
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