# How do you find two consecutive positive integers if the sum of their squares is 3445?

Apr 10, 2018

$\left(42 , 41\right)$

#### Explanation:

First, translate the words into an equation:

${x}^{2} + {y}^{2} = 3445$

Since you know that $x$ and $y$ are consecutive, you know that one of the variables is larger by $1$ compared to the other.

$x = y - 1$

Via substitution, you get the new equation

${x}^{2} + {\left(x - 1\right)}^{2} = 3445$

Expand the second term on the left-hand side

${x}^{2} + {x}^{2} - 2 x + 1 = 3445$

Simplification gets you a quadratic equation $2 {x}^{2} - 2 x - 3444 = 0$.

This can be factored into $2 \left(x - 42\right) \left(x + 41\right)$

By the zero product rule, $x = 42 , x = - 41$

Since they are positive integers, $x = 42$

Because they are consecutive and $x$ is larger than $y$, $y = 41$

The final answer is $\left(42 , 41\right)$