Question

If G.C.D (a, b) = 1, then G.C.D (a + b, a – b) = ?
(a) 1 or 2
(b) a or b
(c) a + b or a – b
(d) 4

Answer 1

Hint: To find the G.C.D, we first need to define what G.C.D is. Then, we will assume some variable for the value of G.C.D (a + b, a – b) and find all the possible values of that variable. It is given that the G.C.D (a, b) = 1. This means that a and b do not have any common factors, we will take help of this condition to find the value of the variable we assumed of the value of G.C.D (a + b, a – b).

Complete step-by-step answer:
The full-form of G.C.D is the greatest common divisor. This means if we are given that G.C.D (x, y) is z, then x and y are divisible by z and there is no other number larger than z which can divide both x and y.
We are given that G.C.D (a, b) = 1. This means that a and b do not have any common divisor, which tells us that a and b, both are prime numbers.
Now, we shall assume that the greatest common divisor of a + b and a – b is d.
This means, both a + b and a – b are divisible by d, i.e. d is a factor of a + b and a – b.
Therefore, we can safely assume that a + b = p(d) and a – b = q(d). Where p and q are other factors of a + b and a – b respectively.
Now, we will add p(d) and q(d).
$\Rightarrow $ 2a = p(d) + q(d)
$\Rightarrow $ 2a = d(p + q)
From the above equation, we can say that d is a factor of 2a.
Similarly, we will now subtract q(d) from p(d).
$\Rightarrow $ 2b = p(d) – q(d)
$\Rightarrow $ 2b = d(p – q)
Thus, we can also say that d is a factor of 2b.
Therefore, this is only possible if either d = 1 or 2.
Hence, G.C.D (a + b, a – b) = 1 or 2.

So, the correct answer is “Option A”.

Note: Another method used to show that the G.C.D (a + b, a – b) = 1 or 2 is based on the nature of a and b. We showed that a and b both are prime numbers. We know that all the prime numbers except 2 are odd. We also know that addition or subtraction of odd numbers give us an even number. Therefore, a + b and a – b, both will be even and thus divisible by 2. If one of a or b is 2, the other is surely an odd number and when 2 is either added or subtracted from that number, it will yield an odd number, which will not be divisible by 2, but will always be divisible by 1.