Question

Find two consecutive positive integers, sum of whose squares is $365$.

Answer 1

Let the first number is $x,$ then the next number will be $x+1.$

Then,

$x^2+({x+1)}^2=365$ {Given}

$\Rightarrow x^2+x^2+2x+1=365$

$\Rightarrow 2x^2+2x-364=0$

$\Rightarrow x^2+x-182=0$

$\Rightarrow x^2+14x-13x-182=0$

$\Rightarrow x(x+14)-13(x+14)=0$

$\Rightarrow (x-13)(x+14)=0$

$\Rightarrow x=13,-14$

Considering positive number, the numbers are $13$ and $14.$