Solving the Equation (a^3b^3 1911ab) for Positive Integers
The problem of finding the positive integer solutions to the equation (a^3b^3 1911ab) involves diving into number theory and algebraic manipulation. This equation can be analyzed through a combination of prime factorization and modular arithmetic. Let's explore the detailed steps to arrive at the solutions for (a) and (b).
Analyzing the Equation
The equation we need to solve is:
[ a^3b^3 1911ab ]First, let's simplify the equation by dividing both sides by (ab), assuming (a) and (b) are positive integers:
[ ab(600^3b^3 - 756^3b^3) 1911ab ]This can be further simplified to:
[ ab(a^3 - b^3) 1911ab ]Note that the simplification should actually be:
[ (a^2b^2)(a^2 - b^2) 1911 ]This simplifies to:
[ (a^2b^2)(a - b)(a b) 1911 ]Prime Factorization of 1911
To further analyze this equation, let's factorize 1911:
[ 1911 3 times 7 times 91 3 times 7^2 times 13 ]The positive integer cubes up to 12 are:
[ 1, 8, 27, 64, 125, 216, 343, 512, 729, 1000, 1331, 1728 ]The relevant cubes involved in the factors of 1911 are:
[ 3^3 27, quad 7^3 343 ]Let's consider the only possible values for (a^3 - b^3):
1. (a^3 - b^3 27) 2. (a^3 - b^3 343)
Case Analysis for (a^3 - b^3)
1. **Case 1: (a^3 - b^3 27)** - This implies (a 3) and (b 0) or (b 3) and (a 0). However, since (a) and (b) are positive integers, this is not a valid solution. - Another possible solution is (a 3) and (b 2), but this does not yield a valid solution for (a^2b^2(a - b)(a b) 1911). - The only valid solution here is (a 3) and (b 4) or (a 4) and (b 3). 2. **Case 2: (a^3 - b^3 343)** - This implies (a 7) and (b 0) or (b 7) and (a 0). Again, since (a) and (b) are positive integers, this is not a valid solution. - The only valid solution here is (a 5) and (b 2), or vice versa. This gives us (a 5) and (b 8) or (a 8) and (b 5).
General Solution Analysis for (a^2b^2(a - b)(a b) 1911)
Let's consider the general case where (a) and (b) have a common factor. We can write (a a_1c_0) and (b b_1c_0), where (a_1) and (b_1) are coprime and (c_0) is a common factor. Substituting this into the original equation, we get:
[ a_1^3b_1^3c_0^3 1911a_1b_1c_0^2 ]Dividing both sides by (a_1b_1c_0^2), we get:
[ a_1^2b_1^2c_0 1911 ]This means (c_0) must be one of the prime factors of 1911, i.e., (c_0 3), (c_0 7), or (c_0 91). For each possible value of (c_0), we need to check if (a_1) and (b_1) can be coprime.
Checking Possible Values for (c_0)
1. **For (c_0 3)** - We need (a_1^2b_1^2 times 3 1911), which gives (a_1^2b_1^2 637). Since 637 is (7^2 times 13), we can have (a_1 7) and (b_1 13), or vice versa. This gives us (a 21) and (b 39), or (a 39) and (b 21). However, these do not satisfy the original equation.
2. **For (c_0 7)** - We need (a_1^2b_1^2 times 7 1911), which gives (a_1^2b_1^2 273). Since 273 is (3 times 7^2), we can have (a_1 3) and (b_1 7), or vice versa. This gives us (a 21) and (b 49), or (a 49) and (b 21). Again, these do not satisfy the original equation.
3. **For (c_0 91)** - We need (a_1^2b_1^2 times 91 1911), which gives (a_1^2b_1^2 21). Since 21 is (3 times 7), we can have (a_1 3) and (b_1 1), or vice versa. This gives us (a 27) and (b 9), or (a 9) and (b 27). Again, these do not satisfy the original equation.
After considering all possible values, the only solutions that satisfy the equation are:
[ a 600, b 960 quad text{and} quad a 756, b 1008 ]Conclusion
The only positive integer solutions to the equation (a^3b^3 1911ab) are:
[ a 600, b 960 quad text{and} quad a 756, b 1008 quad text{or}quad a 960, b 600 quad text{and} quad a 1008, b 756 ]