Question

Question 4
Find two consecutive positive integers, sum of whose squares is 365.

Answer 1

Let the consecutive positive integers be $x$ and $x+1$.
Therefore, $x^2+(x+1{)}^2=365$
$\Rightarrow x^2+x^2+1+2x=365$
$\Rightarrow 2x^2+2x-364=0$
$\Rightarrow x^2+x-182=0$
(Splitting the middle term x as 14x-13x because $14x\times -13x=-182x^2$)
$\Rightarrow x^2+14x-13x-182=0$
$\Rightarrow x(x+14)-13(x+14)=0$
$\Rightarrow (x+14)(x-13)=0$
Either, $x+14=0orx-13=0$,
$\Rightarrow x=-14orx=13$
Since the integers are positive, $x$ can only be $13$.
$\therefore x+1=13+1=14$
Therefore, two consecutive positive integers will be $13$ and $14$.