Question

Find two consecutive natural numbers whose squares have the sum $265$.

Answer 1

Let two consecutive numbers be $x$ and $x+1$

Given:

$x^2+{(x+1)}^2=265$

$x^2+x^2+1+2x=265$

$2x^2+2x-264=0$

$x^2+x-132=0$

$(x+12)(x-11)=0$

$x=11$ and $x=-12$

Hence consecutive numbers are $11,12$ or $-12,-11$