Please help with this problem!!
Assume a, b and c are positive integers such that c cubed=a squared-b squared. If c is the least odd integer for which this is true, what is the least possible value of b?
Caroline
OK, I am working the math questions down from the latest posts, and I notice a lot of questions by you without showing any effort on your part.
Show me what you have so far, and perhaps I can guide you through the problems.
BTW
36 - 9 = 27
27 = 6^2 - 3^2
Previous balance = $179.32
Annual rate = 16%
Finance charge = $ a0
New purchases = $117.42
Payments/credits = $85.00
New balance = $ a1
To solve this problem, we need to understand the concept of perfect squares and perfect cubes.
A perfect square is a number that can be written as the square of an integer. For example, 4, 9, and 16 are perfect squares because they can be written as 2^2, 3^2, and 4^2, respectively.
Similarly, a perfect cube is a number that can be written as the cube of an integer. For example, 8, 27, and 64 are perfect cubes because they can be written as 2^3, 3^3, and 4^3, respectively.
Now, let's proceed with the given problem. We are given that c^3 = a^2 - b^2, and we need to find the least possible value of b.
Since c is the least odd integer for which the given equation is true, we can start by plugging in small odd values for c until we find a valid solution. Let's start with c = 1.
When c = 1, the equation becomes 1^3 = a^2 - b^2. Simplifying it further, we get 1 = a^2 - b^2.
To find the possible values of a and b, we need to factorize the right-hand side of the equation using the difference of squares identity:
a^2 - b^2 = (a + b)(a - b)
From the equation, we have 1 = (a + b)(a - b).
Since a, b, and c are positive integers, we know that both (a + b) and (a - b) must be positive integers. But since we are looking for the least possible value of b, we need to find the smallest values of a and b that satisfy the equation.
Considering a + b = 1 and a - b = 1, we get the following system of equations:
a + b = 1 ...(1)
a - b = 1 ...(2)
Solving this system of equations simultaneously, we find a = 1 and b = 0.
Therefore, the least possible value of b is 0.
Note: In this case, c = 1 and b = 0. However, if you follow the same approach for larger odd values of c, you may find different valid solutions. The solution we obtained here is only for the least odd value of c.