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

Let the 3 numbers be

x-2,x,x+2

(x-2)*x = (x+2)^2 - 34
x=5

So, the numbers are 3,5,7

To solve this problem, we can let the first odd number be x. Since we are looking for three consecutive odd numbers, the next two odd numbers will be x + 2 and x + 4.

According to the problem, the product of the smaller two numbers is 34 less than the square of the largest number. So, the equation to represent this relationship is:

(x)(x + 2) = (x + 4)^2 - 34

Now, let's simplify and solve the equation:

x^2 + 2x = x^2 + 8x + 16 - 34
x^2 + 2x = x^2 + 8x - 18

By subtracting x^2 from both sides of the equation, we get:

2x = 8x - 18

Now, let's isolate x by subtracting 8x from both sides:

2x - 8x = -18
-6x = -18

Divide both sides by -6 to solve for x:

x = -18 / -6
x = 3

So, the first odd number is 3. The next two consecutive odd numbers will be 3 + 2 = 5 and 3 + 4 = 7.

Therefore, the three consecutive odd numbers are 3, 5, and 7.