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

Let the first number be $$x$$

The second number can be represented by $$x + 1$$.

From the given condition,

$$\Rightarrow x^2 + (x + 1)^2 = 365$$
$$\Rightarrow x^2 + x^2 + 2x + 1 - 365 = 0$$
$$\Rightarrow 2x^2 + 2x -364 = 0$$
$$\Rightarrow x^2 + x - 182 = 0$$
(Splitting x as 14x - 13x )
$$\Rightarrow x^2 + 14x - 13x - 182 = 0$$
$$\Rightarrow x(x + 14) - 13(x + 14) = 0$$
$$\Rightarrow (x - 13)(x + 14) = 0$$

The roots of this equation are the values of x for which $$(x + 14)(x - 13) = 0$$

which are,
$$x + 14 = 0$$ or $$x - 13 = 0$$
Thus, $$x = -14$$ or $$x = 13$$

Since the integer is supposed to be positive, $$x = -14$$ is eliminated.

Thus, the two consecutive positive integers are 13 and 13 + 1 = 14.