# There are three consecutive positive integers such that the sum of the squares of the smallest two is 221. What are the numbers?

Jun 22, 2016

There are $10 , 11 , 12$.

#### Explanation:

We can call the first number $n$. The second number has to be consecutive, so it will be $n + 1$ and the third one is $n + 2$.

The condition given here is that the square of the first number ${n}^{2}$ plus the square of the following number ${\left(n + 1\right)}^{2}$ is 221. We can write

${n}^{2} + {\left(n + 1\right)}^{2} = 221$

${n}^{2} + {n}^{2} + 2 n + 1 = 221$

$2 {n}^{2} + 2 n = 220$

${n}^{2} + n = 110$

Now we have two methods to solve this equation. One more mechanics, one more artistic.
The mechanics is to solve the second order equation ${n}^{2} + n - 110 = 0$ applying the formula for the second order equations.
The artistic way is to write

$n \left(n + 1\right) = 110$

and observe that we want that the product of two consecutive numbers has to be $110$. Because the numbers are integer we can search these numbers in the factors of $110$. How can we write $110$?

For example we notice that we can write it as $110 = 10 \cdot 11$.

Oh, it seems we found our consecutive numbers!

$n \left(n + 1\right) = 10 \cdot 11$.

Then $n = 10 , n + 1 = 11$ and, the third number (not very useful for the problem) $n + 2 = 12$.