#### Question

If the greatest common divisor (GCD) of a , b is 1, c, Verify a-b is a perfect square such that \dfrac{ab}{a-b} =c

Question

The greatest common divisor (GCD) of a , b , c is 1, but not necessarily mean GCD of a, b is 1.

Let d be the GCD of a, b, (a > b), then

a = a_1d,

b=b_1d

a-b = (a_1-b_1)d

(1)

in which a_1 and b_1 are prime to each other

c = \dfrac{ab}{a-b}=\dfrac{a_1b_1d^2}{(a_1-b_1)d}=\dfrac{a_1b_1d}{a_1-b_1}

Since a_1 and b_1 are prime to each other,

a_1b_1 and a_1-b_1 are not even or odd numbers simultaneously, that is,

\dfrac{a_1b_1 }{a_1-b_1} is a fraction

In order for c to be an integer, a_1-b_1 must be equal to d

a_1-b_1 = d

Substituting to (1) yields

a-b = d^2