Sum of squares of consecutive integers

Let the smaller integer be $n$ . Then we have:

$\begin{aligned} n^2 + (n+1)^2 & = 365 \\ 2n^2 + 2n + 1 & = 365 \\ n(n+1) & = 182 \\ \implies n & = \left\lfloor \sqrt{182} \right \rfloor = \lfloor 13.4907...\rfloor = \boxed {13} \end{aligned}$

365 ÷ 2 = 182.5
$√(182.5)$ = 13.5092560861, meaning that the smaller of the integers is 13.

x^2 + (x+1)^2 = 365;

2x^2 + 2x + 1 = 365;

2x^2+2x - 364 = 0 or in other words x^2 + x - 182 = 0 , the quadratic equation factorizes as (x+14)(x-13) = 0 so x = -14 or x = 13 , as x > 0 x = 13 is the only solution. Verifying the sum of the squares gives 169 + 196 = 365.