Find all triples of positive integers that satisfy
\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1
Prove that your list is complete (i.e., show there are no other positive integer solutions).
Saen’s Answer
Prerequisites & Existing Solutions
In order to solve this problem, we will first analyse possible sets of three positive integers that will satisfy the equation:
\frac{1}{a}+\frac{1}{b}+\frac{1}{c} = 1
\frac{bc}{abc}+\frac{ac}{abc}+\frac{ab}{abc}=1
bc+ac+ab = abc
Undertaking a simple analysis of the problem shows us a number of solutions that satisfy these criteria. We will denote each individual solution S_n as S_n = [a,b,c]
We will also denote that a \leq b \leq c in each case, as it simplifies the need to rearrange the answers without changing the values.
Solution 1: S_1 = [3,3,3]
We can see logically that adding three thirds together will give us 1/3 + 1/3 +1/3 = 1.
\frac{1}{3}+\frac{1}{3}+\frac{1}{3}
Cross-verification:
bc+ac+ab = abc
(3 \times 3) + (3 \times 3) + (3 \times 3) = (3 \times 3 \times 3)
(9) + (9) + (9) = 27
Solution 2: S_2 = [2,4,4]
Adding one half to one quarter, and then adding another quarter will also give us an answer of 1, and therefore we can see that 1/2 + 1/4 + 1/4 = 1.
Cross-verification:
bc+ac+ab = abc
(4 \times 4) + (2 \times 4) + (2 \times 4) = (2 \times 4 \times 4)
(16) + (8) + (8) = 32
Solution 3: S_3 = [2,3,6]
Adding 1/2 to 1/3 gives us \frac{5}{6}. Adding 1/6 here will give us 1.
Cross-verification:
bc+ac+ab = abc
(3 \times 6) + (2 \times 6) + (2 \times 3) = (2 \times 3 \times 6)
(18) + (12) + (6) = 36
No Other Solutions: Proof by Elimination
There are no other solutions that satisfy the equation. So, how do we prove this empirically?
We will maintain our previous logic of a \leq b \leq c, as once again, it makes no difference to the possible outcomes.
Eliminating a = 0 within solutions
Firstly, it is important to denote that \{a,b,c\} \in \mathbb{Z}^{+}, as per the question requirements. Since 0 is not a positive integer, it will not be included in any valid solution. Inclusion of 0 would also require us to calculate \frac{1}{0}, which would also be impossible.
Eliminating 1 within solutions
Without 0, it becomes impossible to include 1, as the only way to reach the exact value \frac{1}{1} for our answer, where one of the values is 1, would be for a=0, b=1 and c=1 to be used, which is impossible as previously stated.
Eliminating a \geq 4 within solutions
Our test is to verify whether \frac{1}{a}+\frac{1}{b}+\frac{1}{c} = 1. Let it also be noted that in all cases, as the value of x increases, the value of \frac{1}{x} will always decrease. Therefore, for any answer where \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \lt 1, increasing any of a, b, or c would not make a difference, as an increase in the denominator values would always lead to the final value trending further away from 1.
Keeping our a \leq b \leq c logic in mind, if we analyse potential solutions in ascending order such as [4,4,4] being followed by [4,4,5] and then [4,4,6], etc., a proof that [4,4,4] fails our test would also empirically prove that all higher triples, for instance [4,4,5] or [4,4,6], or [5,6,7], as examples, would fail too.
To show that [4,4,4] fails, we can simply input the values:
\frac{1}{4}+\frac{1}{4}+\frac{1}{4} = \frac{3}{4} = 0.75
0.75 \lt 1
therefore, it is impossible for a valid solution to contain an a value that is greater than or equal to 4.
Evaluating our a possibilities following elimination steps
Based on the elimination constraints set out above, it becomes apparent that our only plausible a values are either a=2 or a=3, on the basis that we have eliminated a=1 and any a \geq 4, and \{a \in \mathbb{Z}^+\}. We can now take a look at both cases and determine further elimination steps for b and c.
Evaluation where a=2
An evaluation of
\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 1
where a=2, gives us
\frac{1}{2} + \frac{1}{b} + \frac{1}{c} = 1
\frac{1}{b} + \frac{1}{c} = 1 - \frac{1}{2}
\therefore
\frac{1}{b} + \frac{1}{c} = \frac{1}{2}
\therefore
\frac{1}{\frac{1}{b}+\frac{1}{c}} = \frac{1}{\frac{1}{2}}
\frac{1}{\frac{c}{bc}+\frac{b}{bc}} = 2
\frac{1}{\frac{c+b}{bc}} = 2
\frac{bc}{c+b} = 2
bc = 2(c+b)
In other words, the product of b and c must be equal to double the sum of b and c.
The only positive integer pair that would satisfy this equation in full, is b = 4 and c=4.
To prove that [4,4] is the only positive integer pair where the product is equal to twice the sum, we can assess the relationship xy = 2(x+y), \{x,y \in \mathbb{Z}^+\}, which should only prove true when x and y are both equal to 4.
Since there is no inverse relationship here, and we are dealing only with positive integers, increasing the value of x or y (and subsequently xy) will always lead to a respective increase in the overall value of 2(x+y). Similarly, reducing the value of x or y (and subsequently xy) will always reduce the overall value of 2(x+y).
Let us not forget that the values of xy and 2(x+y) are equivalent when x and y both equal 4. We can now assess [4,4], and the pairs that neighbour it spacewise (excluding repeats) i.e [3,4] to the left and [4,5] to the right.
For [3,4], the 2(x+y) value of 14 is greater than the xy value of 7.
It can therefore be empirically stated that any pairs with a value below [3,4], (such as [3,3], [2,3] and so on) will follow the same pattern, and none will be equal.Similarly, for [4,5], the 2(x+y) value of 18 is less than the xy value of 20. It can therefore be empirically stated that any pairs with a value above [4,5], (such as [5,5], [5,6] and so on) will follow the same pattern, and none will be equal.
Applying both of these logical steps together proves undeniably that [4,4] is the only possible integer pair \in \mathbb{Z}^+ that maintains an equivalence between bc and 2(b+c).
Hence, when a = 2, b must equal 4, and c must equal 4. This means that the only possible solution when a=2 is one where [a,b,c] = [2,4,4].
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Evaluation where a=3
Taking a look at our original equation, \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1, we can see that
\frac{bc}{abc}+\frac{ac}{abc}+\frac{ab}{abc}=1
Substituting 3 for the a-value here gives us
\frac{bc}{3bc}+\frac{3c}{3bc}+\frac{3b}{3bc}=1
\\
bc + 3b + 3c = 3bc
\\
bc + 3(b + c) = 3bc
\\
3(b + c) = 3bc - bc = 2bc
In other words, we now need to show that [3,3] is the only [b,c] pair where twice the product is equivalent to three times the sum.
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Taking a look at spacewise neighbour [2,3], we can see that twice the product, 2(2 \times 3) = 12 is less than three times the sum, 3(2+3)= 15.
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Taking a look at spacewise neighbour [3,4], we can see that twice the product, 2(3 \times 4) = 24 is greater than three times the sum, 3(3+4)= 21.
Similar to before, there are no negative values or inverse relationships in our set. Therefore, it can be stated that [3,3] is the only pair that meets our criteria.
This proves empirically that when a = 3, b and c must also equal 3 for a valid solution.
Final Summary
As shown, there are only three solutions for a, b and c that satisfy the equation
\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1,\{a,b,c \in \mathbb{Z^+}\}
These are the solutions, denoted [a,b,c] where the values are
[2,4,4] \\ [2,3,6] \\ [3,3,3]