How many pairs of positive integers (a,c) satisfy a^3+125=c^3?
A. 0
B. 1
C. 2
D. 3
How Do i figure this out?
I can only see
(0,5) and (-5,0)
the right side has to be a perfect cube
so it could be 1,8,27,64,125, 216
which means that a^3 + 125 must be one of those numbers, and a^3 would be one of
those : 124, 117, 98, ...
does not look too promising other than the 2 I stated
To find the number of pairs of positive integers (a, c) that satisfy the equation a^3 + 125 = c^3, we can start by rearranging the equation to isolate the cubes:
a^3 = c^3 - 125.
Now, notice that 125 can be written as 5^3. So, we can rewrite the equation as:
a^3 = c^3 - 5^3.
Applying the identity a^3 - b^3 = (a - b)(a^2 + ab + b^2), we can factorize the equation as:
(a - c)(a^2 + ac + c^2) = 5^3.
Since 5^3 = 125 and the left-hand side of the equation is the product of two positive integers, we need to find the factor pairs of 125. These factor pairs will give us the corresponding values for (a - c) and (a^2 + ac + c^2). Let's list the factor pairs of 125:
1 × 125
5 × 25
25 × 5
125 × 1
Now, we need to determine which value of the factor pair gives us a valid solution for (a - c) and (a^2 + ac + c^2). Let's analyze each case:
1. (a - c) = 1 and (a^2 + ac + c^2) = 125:
Solving the first equation, we find a = c + 1.
Substituting this value into the second equation, we get:
(c + 1)^2 + (c + 1)c + c^2 = 125.
Simplifying and rearranging, we obtain:
3c^2 + 3c + 3 = 125.
This equation has no positive integer solutions for c.
2. (a - c) = 5 and (a^2 + ac + c^2) = 25:
Solving the first equation, we have a = c + 5.
Substituting this into the second equation, we get:
(c + 5)^2 + (c + 5)c + c^2 = 25.
Simplifying and rearranging, we have:
3c^2 + 11c + 15 = 25.
Rearranging once again, we get:
3c^2 + 11c - 10 = 0.
Solving this quadratic equation, we find c = 1 and c = -10/3. Since c must be a positive integer, c = 1 is the only valid solution. Thus, a = c + 5 = 6.
Therefore, there is one pair of positive integers that satisfies the equation.
3. (a - c) = 25 and (a^2 + ac + c^2) = 5:
Solving the first equation, we obtain a = c + 25.
Substituting this into the second equation, we have:
(c + 25)^2 + (c + 25)c + c^2 = 5.
Simplifying and rearranging, we get:
3c^2 + 51c + 625 = 5.
Rearranging once again, we obtain:
3c^2 + 51c + 620 = 0.
Solving this quadratic equation, we find c = -20/3 and c = -31/3. Since c must be a positive integer, there are no valid solutions for this case.
4. (a - c) = 125 and (a^2 + ac + c^2) = 1:
Solving the first equation, we get a = c + 125.
Substituting this into the second equation, we have:
(c + 125)^2 + (c + 125)c + c^2 = 1.
Simplifying and rearranging, we obtain:
3c^2 + 375c + 15625 = 1.
Rearranging once again, we get:
3c^2 + 375c + 15624 = 0.
Solving this quadratic equation, we find c = -624/3 and c = -12499/3. Since c must be a positive integer, there are no valid solutions for this case.
After analyzing all four cases, we can see that there is only one pair of positive integers (a, c) that satisfies the equation a^3 + 125 = c^3. Therefore, the correct answer is B. 1.