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

1 Answer
Nov 27, 2016

The integers are #9# and #10#

Explanation:

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

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

So:

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

Dividing through by #2# we have:

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

Now #9^2 = 81 < 90.6 < 100 = 10^2#

So:

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

#color(white)()#
Alternatively, we can solve directly:

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

Subtract #181# from both ends to get:

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

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

#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 #n = 9# or #n = -10#

Discard the negative solution since the question specifies positive integers.

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