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

##### 1 Answer

#### Explanation:

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

**Case - positive integer solutions**

Considering positive integers first, let the smallest be denoted by

We are given:

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

which expands to:

#n^2+2n=n+56#

Subtract

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

Which gives us

Since we specified positive integers, the three numbers are

**Case - negative integer solutions**

Now consider negative integers.

If the smallest negative integer is

We are given:

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

which expands to:

#n^2-2n=n+48#

Subtract

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

Multiply by

#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

There are no (negative) integer solutions since