The product of two consecutive odd integers is 1 less than four times their sum. Find the integers.

Let the first odd integer be $x$.
Since the integers are consecutive and odd, the next odd integer will be $x+2$.

The product of the two consecutive odd integers is $x(x+2)$.
Their sum is $x + (x+2) = 2x+2$.
Four times their sum is $4(2x+2)$.
The problem states that the product is 1 less than four times their sum.
So, the equation is:

Expand both sides of the equation:

Rearrange the terms to form a standard quadratic equation $ax^2 + bx + c = 0$:

Factor the quadratic equation:
We need two numbers that multiply to -7 and add to -6. These numbers are -7 and 1.

Set each factor to zero to find the possible values for $x$:

Case 1: If $x = 7$
The first integer is 7.
The second integer is $x+2 = 7+2 = 9$.
The pair of integers is $(7, 9)$.

Case 2: If $x = -1$
The first integer is $-1$.
The second integer is $x+2 = -1+2 = 1$.
The pair of integers is $(-1, 1)$.

Both pairs satisfy the given condition.

The integers are $7 \text{ and } 9 \text{ or } -1 \text{ and } 1$.