Replace every number in the above equation with its reciprocal and the equation is still true. Any equation using the four basic operations that has this property can be rewritten as an equality of two sums that still has this property by using the distributive properties of multiplication and division, applying the multiplications and divisions, and moving subtractions to the other side. (e.g. 3 + 4*(10 – 1) = 30/2 + 24 -> 3 + 4*10 – 4 = 30/2 + 24 -> 3 + 40 – 4 = 15 + 24 -> 3 + 40 = 4 + 15 + 24)
Equations of the form a + b = x + y + z are the simplest possibility that exists non-trivially in the real numbers, and I have found several such equations.
Multiplying all the numbers in the equation by the same number will preserve the property. This scale invariance means that these equations can be put into equivalence classes by scaling. The equivalence class including a + b = x + y + z will be denoted as (a,b|x,y,z). For convenience, I will write numbers on each side in ascending order.
The first six equivalence classes I found that fit within the integers are (3,40|4,15,24), (3,40|5,8,30), (5,45|8,15,24), (8,72|15,24,45), (5,56|7,24,30), and (15,168|28,35,120). There is a simple involution that is very important: (a,b|x,y,z) -> (1/b,1/a|1/z,1/y,1/x).
The way I generate new solutions is quite simple:
Generate three random somewhat-composite numbers1 to go on the right side, then use the formulas S = a + b + c and P = Sabc/(ab+ac+bc) to determine the sum and product of the two numbers on the left side, then (S/2)+-sqrt((S/2)^2-P) to determine the two numbers on the left. Then I have the computer do this over and over until the results are integers or half-integers.
After performing the above search for around an hour, I have come across the solutions (15,760|19,72,684), (55,561|80,264,272), and (48,364|105,112,195).
You may notice that that’s not a particularly useful way of writing these. I’ll write them in prime factorization instead: (3*5, 2*2*2*5*19 | 19, 2*2*2*3*3, 2*2*3*3*19), (5*11, 3*11*17 | 2*2*2*2*5, 2*2*2*3*11, 2*2*2*2*17), and (2*2*2*2*3,2*2*7*13|3*5*7,2*2*2*2*7,3*5*13).
This lets us find their duals in integer form easily – the first has a LCM of 2*2*2*3*3*5*19, the second has an LCM of 2*2*2*2*3*5*11*17, and the third has an LCM of 2*2*2*2*3*5*7*13.
Therefore, their duals are (3*3, 2*2*2*3*19 | 2*5, 5*19, 2*2*2*3*3*5), (2*2*2*2*5, 2*2*2*2*3*17 | 3*5*11, 2*5*17, 3*11*17), and (2*2*3*5,5*7*13|2*2*2*2*7,3*5*13,2*2*2*2*13), which can be written in standard decimal notation as (9,456|10,95,360), (80,816|165,170,561), and (60,455|112,195,208). I also previously found (75,550|152,209,264) (dual: (228,1672|475,600,825)) earlier.
Consider a prime p, and consider the five numbers’ p-adic valuations. Because p-adic valuations are preserved by negation, we can move all the numbers to one side of the bar (e.g. (|-40,-3,4,15,24)).
Now the left side sum’s p-adic valuation is 0 (because the empty sum equals 0). The only way for the sum of two numbers to have a smaller p-adic valuation than either of them is if the two numbers have equal p-adic valuation, so the smallest p-adic valuation in the sum must occur at least twice. Applying this logic to the dual means that the largest p-adic valuation in the sum must also occur at least twice.
If all the p-adic valuations are equal, then you can divide all the numbers by a power of p and get a solution in the integers mod p. For example, the title equation corresponds to (3,5|4,12,3) mod 7, or (3,7|4,4,2) mod 11. I haven’t found any that reduce to (1,1|1,2,2) mod 3 or (1,1|4,4,4) mod 5, despite those being valid in the fields of integers mod 3 and 5. There are none in the field of integers mod 2.
Here’s a proof that there are no self-dual solutions in the rational numbers:
For the sake of contradiction, assume there is such a solution. Then it can be scaled so that the middle of the right side is 1, so it’s: (a/b, b/a|c/d, 1, d/c). This means that a/b + b/a = c/d + 1 + d/c. Then, do the following algebra:
- a/b + b/a = c/d + 1 + d/c
- (a^2+b^2)/ab = (c^2+d^2)/cd + 1
- (a^2+b^2)/ab – (c^2+d^2)/cd = 1
- Note that ab and cd must be equal because of how fractions work
- (a^2+b^2)-(c^2+d^2) = ab = cd
- (a-c)(a+c)-(b-d)(b+d) = (a-d)(a+d)-(b-c)(b+c) = ab = cd
If the value of each side of that final expression is odd, then a, b, c, and d are all odd, but then the first two parts would be even – even = even.
However, if the value of each side of that final expression is even, then one of a and b is even and one of c and d is even. Because a/b and c/d were fractions written in simplest form, only one of a and b is even and only one of c and d is even. Then one of the first two expressions would be… Huh, I thought it would be odd. I guess this isn’t a complete proof after all.
I have not found any way to generate all of these deterministically like Pythagorean triples can. Further research on this topic is needed.
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