Question

The sum of the squares of two consecutive positive integers is 545. Find the sum of those integers.

Answer 1

Let $x$ be one of the positive integers Then the other integer is $x+1$ where $xϵz^{+}$
Since the sum of the squares of the integers is $545$ we get

$x^2+{(x+1)}^2=545$
$\Rightarrow 2x^2+2x-544=0$
$\Rightarrow x^2+x-272=0$
$\Rightarrow x^2+17x-16x-272=0$
$\Rightarrow x(x+17)-16(x+17)=0$
$\Rightarrow (x-16)(x+17)=0$
Here $x=16$ or $x=-17$

But $x$ is a positive integer
Therefore, reject $x=-17$ and take $x=16$

Hence two consecutive positive integers are $16$ and $(16+1)$
i.e., $16$ and $17$