Problem Statement
Find two consecutive positive integers, sum of whose squares is $365$.
Verified Solution & Marking Scheme
Formulate Quadratic Equation
Let the two consecutive positive integers be $x$ and $x + 1$ ($x > 0$):
$$x^2 + (x + 1)^2 = 365$$
$$x^2 + x^2 + 2x + 1 = 365 \implies 2x^2 + 2x - 364 = 0 \implies x^2 + x - 182 = 0$$
Factorize Quadratic
$$x^2 + 14x - 13x - 182 = 0 \implies x(x + 14) - 13(x + 14) = 0$$
$$(x - 13)(x + 14) = 0$$
Since integers are positive, $x \neq -14$. Thus $x = 13$.
State Both Integers
The two integers are $13$ and $13 + 1 = 14$.