The sum of the squares of two consecutive positive integers is 365. Find the integers.

Let the required two consecutive positive integers be x and (x + 1).

According to the given condition,

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

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

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

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

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

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

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

$\Rightarrow x+14=0$ or $x-13=0$

$\Rightarrow x=-14$ or $x=13$

∴ x = 13                (x is a positive integer)

When x = 13,
x + 1 = 13 + 1 = 14

Hence, the required positive integers are 13 and 14.