Question

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

Answer 1

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 `(n+1)^2` is 221. We can write

`n^2+(n+1)^2=221`

`n^2+n^2+2n+1=221`

`2n^2+2n=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(n+1)=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*11`.

Oh, it seems we found our consecutive numbers!

`n(n+1)=10*11`.

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