Question

Let $D=a^2+b^2+c^2$, where $a$ and $b$ are consecutive integers and $c=ab$. Then $\sqrt{D}$ is

• A. always an even integer
• B. always an odd integer
• C. sometimes an odd integer, sometimes not
• D. sometimes a rational number, sometimes not

Answer 1

The correct option is B always an odd integer
$a$ and $b$ are consecutive integers
$b=a+1$ and $c=ab$
Thus,
$D=a^2+(a+1{)}^2+(ab{)}^2$
$=>D=a^2+a^2+2a+1+(ab{)}^2$
$=>D=(ab{)}^2+2a^2+2a+1$
$=>D=(ab{)}^2+2a(a+1)+1$
$=>D=(ab{)}^2+2ab+1$
$=>D=(ab+1{)}^2$
$=>\sqrt{D}=ab+1$ ...........It is odd positive integer because either $a$ or $b$ is even hence $ab$ is even
B