Question

Let $a,b,c$ be positive integer such that $\frac{a\sqrt{2}+b}{b\sqrt{2}+c}$ is a rational number, then which of the following is always an integer?

• A. $\frac{2a^2+b^2}{2b^2+c^2}$
• B. $\frac{a^2+2b^2}{b^2+2c^2}$
• C. $\frac{a^2+b^2-c^2}{a+b+c}$
• D. $\frac{a^2+b^2+c^2}{a+c-b}$

Answer 1

Answer: (D)

$\frac{a^2+b^2+c^2}{a+c-b}$

Rationalising the given term we get

$\left(\frac{a\sqrt{2}+b}{b\sqrt{2}+c}\right)\times \left(\frac{b\sqrt{2}-c}{b\sqrt{2}-c}\right)=\frac{2ab-\sqrt{2}ac+\sqrt{2}{b}^2-bc}{2b^2-c^2}=\frac{b(2a-c)+\sqrt{2}(b^2-ac)}{2b^2-c^2}$

All the numbers are integers so the given term can only be rational if $\sqrt{2}(b^2-ac)$ is rational so, to have the fraction to be integer we should have $b^2-ac=0$

$b^2=ac\Rightarrow b=\sqrt{ac}$

Only option D satisfies this

$=\frac{a^2+b^2+c^2}{a+c-b}=\frac{a^2+c^2+ac}{a+c-\sqrt{ac}}=\frac{{(a+c)}^2-ac}{a+c-\sqrt{ac}}=\frac{(a+c-\sqrt{ac})(a+c+\sqrt{ac})}{a+c-\sqrt{ac}}=a+c+\sqrt{ac}$

$=a+c+b$

which is an integer as $a,b,c$ are all integers