Prove an expression is rational

[anemone]

The rational numbers $a,\,b,\,c$ (for which $a+bc$, $b+ac$ and $a+b$ are all non-zero) satisfy the equality $\dfrac{1}{a+bc} =\dfrac{1}{a+b}-\dfrac{1}{b+ac}$.

Prove that $\sqrt{(c−3)(c+1)}$ is rational.

 

[Opalg]

The rational numbers $a,\,b,\,c$ (for which $a+bc$, $b+ac$ and $a+b$ are all non-zero) satisfy the equality $\dfrac{1}{a+bc} =\dfrac{1}{a+b}-\dfrac{1}{b+ac}$.

Prove that $\sqrt{(c−3)(c+1)}$ is rational.

[sp]Write it as $\dfrac{1}{a+b} = \dfrac{1}{a+bc} + \dfrac{1}{b+ac} = \dfrac{(a+b)(1+c)}{(a+bc)(b+ac)}$. Then $$(a+b)^2(1+c) = (a+bc)(b+ac) = ab(1+c^2) + (a^2+b^2)c,$$ $$(a+b)^2 + 2abc = ab(1+c^2),$$ $$(a+b)^2 = ab(c-1)^2.$$ This shows that $ab = \Bigl(\dfrac{a+b}{c-1}\Bigr)^2$. In other words, $\sqrt{ab} = \Bigl|\dfrac{a+b}{c-1}\Bigr|,$ which is rational.

Also, $c-1 = \dfrac{\pm(a+b)}{\sqrt{ab}}$, so that $c-3 = \dfrac{\pm(a+b)}{\sqrt{ab}} - 2 = \dfrac{\pm(a+b) - 2\sqrt{ab}}{\sqrt{ab}}$, and $c+1 = \dfrac{\pm(a+b)}{\sqrt{ab}} + 2 = \dfrac{\pm(a+b) + 2\sqrt{ab}}{\sqrt{ab}}.$

Therefore $(c-3)(c+1) = \dfrac{(a+b)^2 - 4ab}{ab} = \dfrac{(a-b)^2}{ab}$, from which $\sqrt{(c-3)(c+1)} = \dfrac{|\,a-b\,|}{\sqrt{ab}}$, which is rational.[/sp]