Find two consecutive positive integers,

Solution:

Let the consecutive positive integers be x and x + 1.

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

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

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

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

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

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

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

Either $x+14=0$ or $x-13=0$

i.e., $x=-14$ or $x=13$

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 .