Question

The sum of the squares of three consecutive odd integers is 83. Find the middle odd integer

Answer 1

Let
`2n+1` be the smallest of these integers, so the next two are
`2n+3` and
`2n+5`. The sum of their squares is 83, so

`(2n+1)^2 + (2n+3)^2 + (2n+4)^2 = 83`

Let
`x=2n+3` be the middle number. By substitution, we rewrite the equation as

`(x-2)^2 + x^2 + (x+2)^2 = 83`

Solve for
`x`. Expand the left side.

`(x^2 - 4x + 4) + x^2 + (x^2 + 4x + 4) = 83`

`3x^2 + 8 = 83`

`3x^2 = 75`

`x^2 = 25`

`\implies \boxed{x=\pm5}`