Question

Let $a, b, c$ be positive integers 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. $\frac{2 a^{2}+b^{2}}{2 b^{2}+c^{2}}$
• B. $\frac{a^{2}+b^{2}-c^{2}}{a+b-c}$
• C. $\frac{a^{2}+2 b^{2}}{b^{2}+2 c^{2}}$
• D. $\frac{a^{2}+b^{2}+c^{2}}{a+b-c}$

Answer 1

Answer: (D)

We have, $a, b, c$ are positive integers and $\frac{a \sqrt{2}+b}{b \sqrt{2}+c}$ is a rational number.
$\therefore\left(\frac{a \sqrt{2}+b}{b \sqrt{2}+c}\right)\left(\frac{b \sqrt{2}-c}{b \sqrt{2}-c}\right) $
$=\frac{2 a b-a c \sqrt{2}+b^{2} \sqrt{2}-b c}{2 b^{2}-c^{2}}$
$=\frac{2 a b-b c+\left(b^{2}-a c\right) \sqrt{2}}{2 b^{2}-c^{2}}$
Since $a, b, c$ are positive integers and $\frac{a \sqrt{2}+b}{b \sqrt{2}+c}$ is rational.
$\therefore b^{2}-a c=0$
$\therefore a, b, c$ are in GP.
Let $a=a, b=a r, c=a r^{2}$,
where $r$ is also positive integer.
(a) $\frac{2 a^{2}+b^{2}}{2 b^{2}+c^{2}}$
$=\frac{2 a^{2}+a^{2} r^{2}}{2 a^{2} r^{2}+a^{2} r^{4}}=\frac{1}{r^{2}}$ not integer
(b) $\frac{a^{2}+b^{2}-c^{2}}{a+b-c}=\frac{a^{2}+a^{2} r^{2}-a^{2} r^{4}}{a+a r-a r^{2}}$ $=a\left(\frac{1+r^{2}-r^{4}}{1+r-r^{2}}\right)$ not integer
(c) $\frac{a^{2}+2 b^{2}}{b^{2}+2 c^{2}}$
$=\frac{a^{2}+2 a^{2} r^{2}}{a^{2} r^{2}+2 a^{2} r^{4}}=\frac{1}{r^{2}}$ not integer
(d) $\frac{a^{2}+b^{2}+c^{2}}{a+b-c}=\frac{a^{2}+a^{2} r^{2}+a^{2} r^{4}}{a+a r-a r^{2}}$
$=\frac{a^{2}}{a}\left[\frac{1+r^{2}+r^{4}}{1+r-r^{2}}\right]$
$=a\left(1+r+r^{2}\right)$ integer