Question

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

Answer 1

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`