# How do you find two consecutive positive integers such that the sum of their squares is 181?

Nov 27, 2016

The integers are $9$ and $10$

#### Explanation:

Suppose the consecutive numbers are $n$ and $n + 1$

Then ${n}^{2} < {\left(n + 1\right)}^{2}$

So:

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

Dividing through by $2$ we have:

${n}^{2} < \frac{1}{2} \left({n}^{2} + {\left(n + 1\right)}^{2}\right) = \frac{181}{2} = 90.5 < {\left(n + 1\right)}^{2}$

Now ${9}^{2} = 81 < 90.6 < 100 = {10}^{2}$

So:

$n = 9 \text{ }$ and $\text{ } n + 1 = 10$

$\textcolor{w h i t e}{}$
Alternatively, we can solve directly:

$181 = {n}^{2} + {\left(n + 1\right)}^{2} = {n}^{2} + {n}^{2} + 2 n + 1 = 2 {n}^{2} + 2 n + 1$

Subtract $181$ from both ends to get:

$0 = 2 {n}^{2} + 2 n - 180$

Multiply both sides by $2$ (to make completing the square less painful) to get:

$0 = 4 {n}^{2} + 4 n - 360$

$\textcolor{w h i t e}{0} = 4 {n}^{2} + 4 n + 1 - 361$

$\textcolor{w h i t e}{0} = {\left(2 n + 1\right)}^{2} - {19}^{2}$

$\textcolor{w h i t e}{0} = \left(\left(2 n + 1\right) - 19\right) \left(\left(2 n + 1\right) + 19\right)$

$\textcolor{w h i t e}{0} = \left(2 n - 18\right) \left(2 n + 20\right)$

$\textcolor{w h i t e}{0} = 4 \left(n - 9\right) \left(n + 10\right)$

Hence $n = 9$ or $n = - 10$

Discard the negative solution since the question specifies positive integers.

Hence $n = 9$ and $n + 1 = 10$