Quardratic problem

Let the smaller positive integer be $x$ . Then the larger positive integer is $x+1$ . We have

$\begin{aligned} x^2 + (x+1)^2 &= 313 \nonumber \\ x^2 + x^2 + 2x + 1 &= 313 \nonumber \\ 2x^2 + 2x - 312 &= 0 \nonumber \\ x^2 + x - 156 &= 0 \nonumber \\ (x+13)(x-12) &=0 \nonumber \end{aligned}$

Since $x$ is positive, $x=12$ . Thus, the sum of the two integers is $12+13=\boxed{25}$ .

Let the first positive integer be $x$ .

$\implies$ The second positive integer is $x+1$ .

$x^2 + (x+1)^2 = 313$ $2x^2 +2x + 1 = 313$ $x^2+x-156 = 0$ $(x-12)(x+13)=0$ $x=12 \quad \text{or} \quad -13 \quad \text{(rej.)}$

Therefore, sum $= 12+12+1 = \boxed{25}$ .

x^2 + y^2 = 313 , y=x+1 , x^2 +x^2 +1 +2x = 313, , 2x^2 +2x-312 = 0, , x^2 +x -156= 0 , x = 12 , y = 13 , 12+13 = 25 Here is your answer