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