Integer Solutions To 1/a + 1/b + 1/c + 1/d = 1

by Ahmed Latif 47 views

Hey guys! Let's dive into a fascinating problem today: 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 condition that a < b < c < d. This is a classic problem in Diophantine equations, and it's a real brain-teaser! We'll break it down step by step, so you can follow along and understand the solution process.

Understanding the Problem

Before we jump into solving, let's make sure we fully grasp what the question is asking. We need to find sets of four distinct positive integers (a, b, c, and d) where the sum of their reciprocals equals 1. The catch? The integers must be in strictly increasing order (a < b < c < d). This condition is crucial because it limits the possible solutions and helps us avoid counting the same solution multiple times with different orderings.

Diophantine equations, like this one, are polynomial equations where we're only interested in integer solutions. They often appear in number theory and have a rich history. This particular problem combines the Diophantine equation aspect with an inequality constraint, making it a fun challenge.

The core of solving Diophantine equations often lies in clever manipulation, bounding the variables, and systematically checking possibilities. There isn't one single method that works for all Diophantine equations, so we need to adapt our approach based on the specific equation's structure. For this problem, we will use a combination of logical deduction and casework.

Let's think about the smallest possible value for 'a'. If 'a' were greater than or equal to 4, then even if b, c, and d were the smallest possible integers greater than 'a' (i.e., a+1, a+2, a+3), the sum of the reciprocals would be less than 1. This is a crucial observation because it gives us a starting point. We know that 'a' must be less than 4. This significantly narrows down our search.

So, we'll consider the possible values for 'a' one by one (1, 2, and 3) and see what restrictions each value puts on b, c, and d. For each value of 'a', we'll simplify the equation and try to find possible values for 'b'. Then, for each pair of (a, b), we'll look for possible (c, d) pairs. This systematic approach will help us ensure we don't miss any solutions.

It's like a puzzle, guys! We have a few pieces of information, and we need to fit them together in a logical way to reveal the complete picture. Don't worry if it seems a bit daunting at first. We'll take it step by step, and you'll see how it all comes together.

Case 1: a = 2

Let's start with the case where a = 2. Our equation now becomes:

1/2 + 1/b + 1/c + 1/d = 1

Subtracting 1/2 from both sides, we get:

1/b + 1/c + 1/d = 1/2

Now, since b > a, we know that b > 2. If b were greater than or equal to 6, then the maximum value of 1/c + 1/d (even if c and d were the smallest possible integers greater than b) would be less than 1/2. This means b must be less than 6. So, the possible values for b are 3, 4, and 5. We'll explore each of these possibilities.

Subcase 1: b = 3

If b = 3, our equation becomes:

1/3 + 1/c + 1/d = 1/2

Subtracting 1/3 from both sides:

1/c + 1/d = 1/2 - 1/3 = 1/6

Since c > b, we know c > 3. If c were greater than or equal to 13, then the maximum value of 1/d would be too small to make the sum equal to 1/6. So, c must be less than 13. Let's explore the possible values of c:

  • If c = 7, then 1/d = 1/6 - 1/7 = 1/42, so d = 42. This gives us the solution (2, 3, 7, 42).
  • If c = 8, then 1/d = 1/6 - 1/8 = 1/24, so d = 24. This gives us the solution (2, 3, 8, 24).
  • If c = 9, then 1/d = 1/6 - 1/9 = 1/18, so d = 18. This gives us the solution (2, 3, 9, 18).
  • If c = 10, then 1/d = 1/6 - 1/10 = 1/15, so d = 15. This gives us the solution (2, 3, 10, 15).
  • If c = 11, then 1/d = 1/6 - 1/11 = 5/66, so d = 66/5, which is not an integer. So, no solution here.
  • If c = 12, then 1/d = 1/6 - 1/12 = 1/12, so d = 12. But this is not allowed since c < d. No solution here.

Subcase 2: b = 4

If b = 4, the equation becomes:

1/4 + 1/c + 1/d = 1/2

Subtracting 1/4 from both sides:

1/c + 1/d = 1/2 - 1/4 = 1/4

Since c > b, we have c > 4. If c were greater than or equal to 9, the maximum value of 1/d would be too small. So, c must be less than 9. Let's check the possibilities:

  • If c = 5, then 1/d = 1/4 - 1/5 = 1/20, so d = 20. This gives us the solution (2, 4, 5, 20).
  • If c = 6, then 1/d = 1/4 - 1/6 = 1/12, so d = 12. This gives us the solution (2, 4, 6, 12).
  • If c = 7, then 1/d = 1/4 - 1/7 = 3/28, so d = 28/3, which is not an integer. No solution here.
  • If c = 8, then 1/d = 1/4 - 1/8 = 1/8, so d = 8. But this is not allowed since c < d. No solution here.

Subcase 3: b = 5

If b = 5, our equation becomes:

1/5 + 1/c + 1/d = 1/2

Subtracting 1/5 from both sides:

1/c + 1/d = 1/2 - 1/5 = 3/10

Since c > b, we have c > 5. If c were greater than or equal to 7, the maximum value of 1/d would be too small. So, c must be less than 7. Let's see the possibilities:

  • If c = 6, then 1/d = 3/10 - 1/6 = 4/60 = 1/15, so d = 15. This gives us the solution (2, 5, 6, 15).

So, from Case 1 (a = 2), we have found a total of 7 solutions.

Case 2: a = 3

Now, let's consider the case where a = 3. Our equation now looks like this:

1/3 + 1/b + 1/c + 1/d = 1

Subtracting 1/3 from both sides, we get:

1/b + 1/c + 1/d = 2/3

Since b > a, we know b > 3. If b were greater than or equal to 4, let's think about the possibilities. We need 1/b + 1/c + 1/d to equal 2/3. The smallest possible value for b is 4. Let's explore that first.

If b = 4, then:

1/4 + 1/c + 1/d = 2/3

Subtracting 1/4 from both sides:

1/c + 1/d = 2/3 - 1/4 = 5/12

Now, c > b, so c > 4. If c were greater than or equal to 3, the sum 1/c + 1/d could never reach 5/12. Checking values for 'c':

  • If c = 3 is not possible because the value must be grater than b
  • If c=4 is not possible because the value must be grater than b
  • If c = 5, then 1/d = 5/12 - 1/5 = 13/60, so d = 60/13, which is not an integer. No solution here.
  • If c = 6, then 1/d = 5/12 - 1/6 = 1/4, so d = 4. But this is not allowed since c < d. No solution here.

So, no solutions for the case b=4. Let's consider the case b > 4. If b were greater than or equal to 3, the sum 1/b + 1/c + 1/d could never be great than 2/3.

So, from Case 2 (a = 3), we have found a no solution.

Case 3: a = 1

Finally, let's look at the a = 1 case. Our equation is:

1/1 + 1/b + 1/c + 1/d = 1

Which simplifies to:

1 + 1/b + 1/c + 1/d = 1

Subtracting 1 from both sides gives:

1/b + 1/c + 1/d = 0

Since b, c, and d are positive integers, their reciprocals will also be positive. The sum of three positive numbers cannot be zero. Therefore, there are no solutions for the case where a = 1.

The Final Count

Alright, guys! We've systematically gone through all the possible cases. Let's recap:

  • Case 1 (a = 2): We found 7 solutions.
  • Case 2 (a = 3): We found 0 solutions.
  • Case 3 (a = 1): We found 0 solutions.

So, in total, there are 7 positive integral solutions (a, b, c, d) that satisfy the given equation and condition.

It was a bit of a journey, but we made it! This problem beautifully illustrates how breaking down a complex problem into smaller, manageable cases can lead us to the solution. Remember the key techniques we used here: bounding the variables, casework, and logical deduction. These are valuable tools in your problem-solving arsenal! Keep practicing, and you'll become a Diophantine equation-solving master in no time! Hura!