Find three consecutive odd integers such that the first integer times thesecond integer is one less than four times the third

Let's denote the three consecutive odd integers as x, x+2, and x+4.

According to the problem, the first integer times the second integer is one less than four times the third. This can be written as:

x*(x+2) = 4*(x+4) - 1

Expanding and simplifying the equation gives:

x^2 + 2x = 4x + 16 - 1

x^2 + 2x - 4x - 15 = 0

x^2 - 2x - 15 = 0

This is a quadratic equation in the form ax^2 + bx + c = 0. We can solve it by factoring:

(x - 5)(x + 3) = 0

Setting each factor equal to zero gives the solutions x = 5 and x = -3.

However, since we're looking for positive odd integers, we discard x = -3.

So, the three consecutive odd integers are 5, 7, and 9.