Question

if HCF of a and b is equal to 1 then why can't 3 be common divisor of a and b

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Answer 1

Answer:

Because if 3 is a common divisor of a and b then, 3 will divide both a and b, as HCF means the greatest number that will divide all the given numbers. Here HCF is 1, which indicates only 1 divides both a and b. Hence 3 cannot be a common divisor.

done

Answer 2

Step-by-step explanation:

It is given that GCD(a,b)=1

Let GCD(a−b,a+b)=d

⇒d divides a−b and a+b

there exists integers m and n such that 

a+b=m×d        ..........(1)

and a−b=n×d        ..........(2)

Upon adding and subtracting equation (1) and (2) we get

2a=(m+n)×d         ..........(3)

and 2b=(m−n)×d         ..........(4)

Since, GCD(a,b)=1(given)

∴2×GCD(a,b)=2

∴GCD(2a,2b)=2 since GCD(ka,kb)=kGCD(a,b)

Upon substituting  value of 2a and 2b from equations (3) and (4) we get

∴gcd((m+n)×d,(m−n)×d)=2

∴d×gcd((m+n),(m−n))=2

∴d× some integer=2

∴d divides 2

∴d≤2 if x divides y, then ∣x∣≤∣y∣

∴d=1 or 2 since, gcd is always a positive integer.