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

##### 1 Answer

The integers are

#### Explanation:

Suppose the consecutive numbers are

Then

So:

#2n^2 < n^2 + (n+1)^2 < 2(n+1)^2#

Dividing through by

#n^2 < 1/2(n^2+(n+1)^2) = 181/2 = 90.5 < (n+1)^2#

Now

So:

#n = 9" "# and#" "n + 1 = 10#

Alternatively, we can solve directly:

#181 = n^2+(n+1)^2 = n^2+n^2+2n+1 = 2n^2+2n+1#

Subtract

#0 = 2n^2+2n-180#

Multiply both sides by

#0 = 4n^2+4n-360#

#color(white)(0) = 4n^2+4n+1-361#

#color(white)(0) = (2n+1)^2-19^2#

#color(white)(0) = ((2n+1)-19)((2n+1)+19)#

#color(white)(0) = (2n-18)(2n+20)#

#color(white)(0) = 4(n-9)(n+10)#

Hence

Discard the negative solution since the question specifies positive integers.

Hence