Question

Two consecutive positive integers, sum of whose squares is $365$ are

• A. $13,14$
• B. $14,15$
• C. $12,13$
• D. $11,12$

Answer 1

The correct option is A $13,14$

Let the two consecutive positive integers be $x$ and $x+1$
Then,
$x^2+(x+1{)}^2=365$
$\Rightarrow x^2+x^2+2x+1=365$
$\Rightarrow 2x^2+2x-364=0$
$\Rightarrow x^2+x-182=0$
Using the quadratic formula, we get
$x=\frac{-1\pm \sqrt{1+728}}{2}$
$\Rightarrow \frac{-1\pm 27}{2}$
$\Rightarrow x=13$ and $x=-14$
But $x$ is given to be a positive integer. $\therefore x\ne -14$
Hence, the two consecutive positive integers are $13$ and $14$.