# The square of the first added to twice the second is 5, what are the two integers?

Jun 27, 2015

There are an infinite number of solutions, the simplest and only positive integer solutions being 1 and 2.

#### Explanation:

For any $k \in \mathbb{Z}$

let $m = 2 k + 1$
and $n = 2 - 2 k - 2 {k}^{2}$

Then:

${m}^{2} + 2 n$

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

$= 4 {k}^{2} + 4 k + 1 + 4 - 4 k - 4 {k}^{2} = 5$

Jun 27, 2015

If they are supposed to be consecutive integers, then the solution with negatives is the first is $- 3$ and the second is $- 2$.
The positive solution is: first is $1$ and second is $2$.

#### Explanation:

Assuming that these are supposed to be consecutive integers and the lesser integer is the first, then we can use:

first = $n$ and second = $n + 1$

The square of the first is ${n}^{2}$ and twicwe the second is $2 \left(n + 1\right)$, so we get the equation:

${n}^{2} + 2 \left(n + 1\right) = 5$

(Note that this is not a linear equation. It is quadratic.)

Solve:

${n}^{2} + 2 \left(n + 1\right) = 5$
${n}^{2} + 2 n + 2 = 5$
${n}^{2} + 2 n - 3 = 0$
$\left(n + 3\right) \left(n - 1\right) = 0$

$n + 3 = 0$ leads to $n = - 3$ and $n + 1$ = -2

If we check the answer, we get ${\left(- 3\right)}^{2} + 2 \left(- 2\right) = 9 + \left(- 4\right) = 5$

$n - 1 = 0$ leads to $n = 1$ and $n + 1$ = 2

If we check this answer, we get ${\left(1\right)}^{2} + 2 \left(2\right) = 1 + 4 = 5$