find 3 consecutive odd numbers where the product of the smaller two number is 34 less than the square of the largest number

Let the three numbers be 2n-1, 2n + 1 and 2n+3  ......so we have

(2n - 1)(2n + 1) + 34 = (2n + 3)^2    simplify

4n^2 - 1 + 34 = 4n^2 + 12n + 9      

33 = 12n  + 9

24 = 12n     divide by 12 on both sides

2 = n

The first odd number is 2(2) - 1 = 3

And the next two are 5 and 7