How do you find three consecutive odd integers such that the product of the two smaller exceeds the largest by 52?

Note the question speaks of "smaller" and "largest" numbers rather than "lesser" and "greatest", so it is easier to split positive and negative cases...

`color(white)()`
Case - positive integer solutions

Considering positive integers first, let the smallest be denoted by `n`, so the other two are `n+2` and `n+4`.

We are given:

`n(n+2) = (n+4)+52`

which expands to:

`n^2+2n=n+56`

Subtract `n+56` from both sides to get:

`0 = n^2+n-56 = (n+8)(n-7)`

Which gives us `n=7` or `n=-8`.

Since we specified positive integers, the three numbers are `7, 9, 11`

`color(white)()`
Case - negative integer solutions

Now consider negative integers.

If the smallest negative integer is `n`, then the other two are `n-2` and `n-4`.

We are given:

`n(n-2)=(n-4)+52`

which expands to:

`n^2-2n=n+48`

Subtract `n+48` from both sides to get:

`0 = n^2-3n-48`

Multiply by `4` so we can complete the square with integers:

`0 = 4(n^2-3n-48)`

`color(white)(0) = 4n^2-12n-192`

`color(white)(0) = (2n)^2-2(2n)(3)+9-201`

`color(white)(0) = (2n+3)^2-201" "` (I could stop here, but...)

`color(white)(0) = (2n+3)^2-(sqrt(201))^2`

`color(white)(0) = ((2n+3)-sqrt(201))((2n+3)+sqrt(201))`

`color(white)(0) = (2n+3-sqrt(201))(2n+3+sqrt(201))`

`color(white)(0) = 4(n+3/2-sqrt(201)/2)(n+3/2+sqrt(201)/2)`

So `n = -3/2+-sqrt(201)/2" "` (not an integer)

There are no (negative) integer solutions since `201` is not a perfect square.