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

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

Uniteasy Admin · Accepted Answer

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
$$a = a_1d$$, 
$$b=b_1d$$
$$a-b = (a_1-b_1)d$$n
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
$$a-b = d^2 $$

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