Solve 1/a+1/b+1/c+1/d=1: Integer Solutions Explained

Hey guys! Today, we're diving into a fascinating problem from the realm of Diophantine equations: finding the number of positive integer solutions (a, b, c, d) that satisfy the equation 1/a + 1/b + 1/c + 1/d = 1, with the added condition that a < b < c < d. This means we're looking for unique combinations of four different positive integers whose reciprocals add up to 1. This question often pops up in math competitions and is a great exercise in number theory and problem-solving techniques. Let's break it down step-by-step and explore how to tackle this kind of problem. We'll explore the constraints, systematically narrow down the possibilities, and ultimately count the solutions. So, buckle up and let's get started!

Understanding the Constraints and Initial Approach

In approaching Diophantine equations , especially one like this, the initial key is recognizing and understanding the constraints imposed on the variables. Here, we have several constraints. First, a, b, c, and d must be positive integers. This immediately restricts our search space from the infinite realm of real numbers to the much smaller, but still substantial, world of integers. Second, the condition a < b < c < d is crucial. This tells us that the integers are distinct and ordered, which helps us avoid overcounting solutions that are simply permutations of the same numbers. This ordering also implies that 'a' is the smallest integer, and 'd' is the largest.

Considering these constraints, let's focus on 'a' first. Since a < b < c < d, we know that 1/a must be the largest fraction among the four. To satisfy the equation 1/a + 1/b + 1/c + 1/d = 1, 1/a must be less than 1 but also leave enough room for the other three fractions to add up to the remainder. If 'a' were 1, then 1/a would be 1, leaving no room for 1/b, 1/c, and 1/d. Therefore, 'a' must be greater than 1. Also, if 'a' were greater or equal to 4, then the maximum value for the left side of the equation would be 1/4 + 1/5 + 1/6 + 1/7 which is less than 1. Therefore, a must be less than 4. Therefore, the possible values for 'a' are 2 and 3. This is a crucial first step: narrowing down the possibilities for 'a' significantly restricts the search space for the other variables. Next, we'll consider each possible value of 'a' separately and explore the implications for b, c, and d.

Considering a = 2*, we have the equation 1/2 + 1/b + 1/c + 1/d = 1. This simplifies to 1/b + 1/c + 1/d = 1/2. Now, we need to find integers b, c, and d such that 2 < b < c < d and their reciprocals sum to 1/2. The same logic applies here: 1/b must be the largest fraction among the remaining three. If 'b' is equal or greater than 6, the maximum sum of the fractions 1/b + 1/c + 1/d would be 1/6 + 1/7 + 1/8 which is less than 1/2. Thus, 'b' must be less than 6. Since b > a = 2, the possible values for 'b' are 3, 4, and 5. We'll analyze each of these cases separately.

If b = 3, our equation becomes 1/c + 1/d = 1/2 - 1/3 = 1/6. Now we seek integers c and d such that 3 < c < d and 1/c + 1/d = 1/6. We know that 'c' must be greater than 6 (because if c is equal or less than 6, then 1/c >= 1/6, and 1/d would be zero or negative). Also, 1/c must be less than 1/6, meaning c > 6. Furthermore, if c >= 13, then 1/c + 1/d <= 1/13 + 1/14 which is less than 1/6. Therefore, c must be less than 13. Thus, the possible values for 'c' are 7, 8, 9, 10, 11, and 12. For each value of 'c', we can find the corresponding 'd' using the equation d = 1 / (1/6 - 1/c). We will then check if 'd' is an integer and if c < d. This gives us the solutions (7, 42), (8, 24), (9, 18), (10, 15), and (12, 12). However, we must exclude (12,12) since c must be less than d. So, we have 4 solutions for this case.

If b = 4, the equation becomes 1/c + 1/d = 1/2 - 1/4 = 1/4. Following a similar approach, 'c' must be greater than 4, and since 1/c must be less than 1/4, c > 4. Also, if c >= 9, then 1/c + 1/d <= 1/9 + 1/10 which is less than 1/4. Therefore, c must be less than 9. Thus, the possible values for 'c' are 5, 6, 7, and 8. Again, we find 'd' using d = 1 / (1/4 - 1/c) and check the conditions. This yields the solutions (5, 20), (6, 12), and (8, 8). We exclude (8,8) since c < d. So, we have 2 solutions for this case.

If b = 5, the equation is 1/c + 1/d = 1/2 - 1/5 = 3/10. Here, 'c' must be greater than 5, and 1/c < 3/10, implying c > 10/3 ≈ 3.33. Also, if c >= 6, then 1/c + 1/d <= 1/6 + 1/7 which is less than 3/10. Since c must be greater than 5, there are no possible values for c. Thus, there are no solutions for this case.

Considering a = 3, we have 1/3 + 1/b + 1/c + 1/d = 1, which simplifies to 1/b + 1/c + 1/d = 2/3. Since 3 < b < c < d, we know that b must be greater than 3. If b >= 4, then 1/b <= 1/4, and the sum of 1/b + 1/c + 1/d would have a maximum value when b, c, and d are as small as possible. If b is equal or greater than 3, the maximum value of the left side would be 1/3 + 1/4 + 1/5, which is less than 2/3. Thus, 'b' must be less than 3. The value of b must be greater than a, so b cannot be equal to a = 3. Therefore, the possible values for 'b' are 2. This cannot happen because we know a < b.

If b = 2, this contradicts our initial condition that a < b, since we're now considering the case where a = 3. This scenario is not valid under the given constraints.

If b = 3, since we already established that a = 3, this violates the condition a < b, making this case invalid as well.

Let's correct the approach for when a = 3. We have the equation 1/3 + 1/b + 1/c + 1/d = 1, which simplifies to 1/b + 1/c + 1/d = 2/3. Since 3 < b < c < d, we need to find suitable values for b, c, and d. We know b must be greater than 3. If b is equal or greater than 3, the maximum value of the left side would be 1/3 + 1/4 + 1/5 which is less than 2/3. Thus, 'b' must be less than 3. But since 'b' needs to be greater than a = 3, this is a contradiction. Therefore, b must be between 3 and some upper bound that makes the sum 2/3 achievable. Let's analyze the possible values for b:

If b = 2, this contradicts the given condition a < b, as a = 3 in this case. Thus, b cannot be 2. If b = 3, this again violates the condition a < b, making this case invalid as well. Let's reconsider the constraints. Since a = 3, we have 1/b + 1/c + 1/d = 2/3, with 3 < b < c < d. The smallest possible value for b is 2, but this contradicts the condition a < b. Therefore, let's analyze the minimum values for 1/b, 1/c, and 1/d to ensure they sum up to 2/3. Since b > 3, let's start with b = 2.

If b = 2: The equation 1/b + 1/c + 1/d = 2/3 can't hold if a = 3 and a < b because b cannot be less than a. If b = 3: Here, 1/3 + 1/c + 1/d = 2/3, which simplifies to 1/c + 1/d = 1/3. Since c > b = 3, c must be at least 4. Also, if c >= 7, then 1/c + 1/d would be at most 1/7 + 1/8, which is less than 1/3. Therefore, c must be less than 7. The possible values for c are 4, 5, and 6. If c = 4, then 1/d = 1/3 - 1/4 = 1/12, so d = 12. This gives us the solution (3, 4, 12). If c = 5, then 1/d = 1/3 - 1/5 = 2/15, so d = 15/2, which is not an integer. If c = 6, then 1/d = 1/3 - 1/6 = 1/6, so d = 6. But c < d must hold, so this solution is not valid.

Counting the Solutions

Summing up the solutions, we found 4 solutions when a = 2 and b = 3, 2 solutions when a = 2 and b = 4, and 1 solution when a = 3 and b = 4. Therefore, the total number of positive integer solutions (a, b, c, d) satisfying the given conditions is 4 + 2 + 1 = 7.

Therefore, there are a total of 6 + 1 = 7 solutions. These solutions are:

(2, 3, 7, 42) (2, 3, 8, 24) (2, 3, 9, 18) (2, 3, 10, 15) (2, 4, 5, 20) (2, 4, 6, 12) (3, 3, 4, 12)

Therefore, there are 7 sets of positive integers (a, b, c, d) satisfying the equation 1/a + 1/b + 1/c + 1/d = 1 with a < b < c < d.

Conclusion

In conclusion, finding the number of positive integral solutions for Diophantine equations requires a systematic approach. By understanding and applying the constraints, narrowing down possibilities, and carefully checking each case, we can successfully solve these problems. This particular problem showcased how bounding the variables and considering cases can lead to a complete solution. Remember, the key is to be organized, patient, and persistent. Keep practicing, and you'll become a pro at solving these types of problems!