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

(A) Let the consecutive positive integers be $x$ and $x+1$.
Given that $x^{2} + (x+1)^{2} = 365$.
Expanding the equation: $x^{2} + x^{2} + 2x + 1 = 365$.
Simplifying: $2x^{2} + 2x + 1 = 365$.
$2x^{2} + 2x - 364 = 0$.
Dividing by $2$: $x^{2} + x - 182 = 0$.
Factoring the quadratic equation: $x^{2} + 14x - 13x - 182 = 0$.
$x(x + 14) - 13(x + 14) = 0$.
$(x + 14)(x - 13) = 0$.
This gives $x = -14$ or $x = 13$.
Since the integers must be positive,we take $x = 13$.
Therefore,the next integer is $x + 1 = 13 + 1 = 14$.
The two consecutive positive integers are $13$ and $14$.