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

1 Answer
Burglar
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 #(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#.

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