Find the values of x , y , z and t

[Kodafafabe]

Hello every one!
This is my first post and it's a great honor to join you on this forum.

X, y , z and t are naturals whole .
(1/x)+(1/y)+(1/z)+(1/t)=1/3.

J'ai montré que (x+y)/xy + (z+t)/zt = 1/3

 

[blamocur]

This equation has lots of solutions. Are there other conditions/restrictions?

 

[Otis]

x , y , z and t are [Whole numbers]

(1/x) + (1/y) + (1/z) + (1/t) = 1/3

Find the values of x , y , z and t

Hi. Welcome to the boards.

The equation above has many solutions. Which solution are you looking for?

[I showed that] (x+y)/(xy) + (z+t)/(zt) = 1/3

That's true. Is that construction a part of the exercise?

PS: You need to type grouping symbols around numerators and denominators, whenever they consist of more than a single number.

 

[Kodafafabe]

Thanks.
Oki i understand about the " many solutions" that's why I need a an expression that can help me finding all the values.

That construction is equivalent to the first expression

 

[Otis]

I need [an] expression that can help me [find] all [solutions].

Why are you interested in all solutions, Kodafafabe?

You have but one equation relating four unknowns. That's not enough information for any algebraic method.

You need an algorithm, to generate solutions. Perhaps later, after analyzing results, numerical patterns might be found for generating some solution subsets. Otherwise, a computer program (using nested loops) is easiest. It will list all solutions within thresholds that you choose.

Also, you may have already realized that, once you obtain a solution, the Commutative Property tells us that permutations are also solutions. For example, if you were to discover four, distinct x,y,z,t values (eg: 6,11,22,33), then all 24 permutations are solutions.

Earlier, I'd realized that 1/3 = 4/12, so one x,y,z,t solution is 1/12, 1/12, 1/12, 1/12. A computer program confirms that no other solutions exist, unless one or more variables become larger than 12. For example, if we allow values up to 15, then there are nineteen solutions:

10, 10, 15, 15
10, 12, 12, 15
10, 12, 15, 12
10, 15, 10, 15
10, 15, 12, 12
10, 15, 15, 10
12, 10, 12, 15
12, 10, 15, 12
12, 12, 10, 15
12, 12, 12, 12
12, 12, 15, 10
12, 15, 10, 12
12, 15, 12, 10
15, 10, 10, 15
15, 10, 12, 12
15, 10, 15, 10
15, 12, 10, 12
15, 12, 12, 10
15, 15, 10, 10

Setting the upper threshold at 100 results in more than 2,200 solutions.

I'm not sure what else to suggest, without knowing your end goal.

 

[JeffM]

This is a slight variation on the previous answers.

In general, if you have n unknowns, you need n equations that are INDEPENDENT, CONSISTENT, and LINEAR in order to find unique values for each unknown. It is true that if you impose the extra constraint that solutions must be integers, you sometimes may need fewer than n equations, but that is not necessarily true.

In this case, you have four unknowns and one equation that is not linear. Why would you even expect to find a unique solution simply by imposing the extra constraint that the solution must consist of positive integers?

 

[blamocur]

I cannot prove it, but my computer believes that the 339 quadruplets listed in the attached files are all possible solutions to that equation. I would be very interested to hear if anyone can either find other solutions or can prove that none exist.

 

[JeffM]

I cannot prove it, but my computer believes that the 339 quadruplets listed in the attached files are all possible solutions to that equation. I would be very interested to hear if anyone can either find other solutions or can prove that none exist.

[math]m,\ n, \ p \in \mathbb Z^+ \text { and } \dfrac{1}{m} + \dfrac{1}{n} = \dfrac{1}{p} \implies p(n + m) = mn.[/math]
Therefore mn is a multiple of p. Without loss of generality, we can say m is a multiple of p.

[math]\therefore p(m + n) = mn \implies p(kp + n) = kpn \implies kp + n = kn \implies p = \dfrac{nk - n}{k} = n - \dfrac{n}{k}.[/math]
But that means n/k is an integer, which means there exists integer j such that n = jk.

[math]\therefore p = kj - j = j(k - 1).[/math]
So in this much simpler case, we can see that there are no solutions if p is prime and a finite number of solutions if p is composite.

For example if p = 12, we could have j = 3 and k = 5. Then n = 15. And m = 60.

[math]\dfrac{1}{15} + \dfrac{1}{60} = \dfrac{5}{60} = \dfrac{1}{12}.[/math]
Or we could have j = 2 and k = 7.

So n = 14. And m = 84.

[math]\dfrac{1}{14} + \dfrac{1}{84} = \dfrac{6}{84} + \dfrac{1}{84} = \dfrac{7}{84} = \dfrac{1}{12}[/math]
Can this be extended by induction?

 

[blamocur]

[math]m,\ n, \ p \in \mathbb Z^+ \text { and } \dfrac{1}{m} + \dfrac{1}{n} = \dfrac{1}{p} \implies p(n + m) = mn.[/math]
Therefore mn is a multiple of p. Without loss of generality, we can say m is a multiple of p.

[math]\therefore p(m + n) = mn \implies p(kp + n) = kpn \implies kp + n = kn \implies p = \dfrac{nk - n}{k} = n - \dfrac{n}{k}.[/math]
But that means n/k is an integer, which means there exists integer j such that n = jk.

[math]\therefore p = kj - j = j(k - 1).[/math]
So in this much simpler case, we can see that there are no solutions if p is prime and a finite number of solutions if p is composite.

For example if p = 12, we could have j = 3 and k = 5. Then n = 15. And m = 60.

[math]\dfrac{1}{15} + \dfrac{1}{60} = \dfrac{5}{60} = \dfrac{1}{12}.[/math]
Or we could have j = 2 and k = 7.

So n = 14. And m = 84.

[math]\dfrac{1}{14} + \dfrac{1}{84} = \dfrac{6}{84} + \dfrac{1}{84} = \dfrac{7}{84} = \dfrac{1}{12}[/math]
Can this be extended by induction?

Using your approach it is not too difficult to prove that in every solution at least one number is divisible by 3. But I don't see what else can be done to prove that the number of solutions is finite.

 

[JeffM]

Using your approach it is not too difficult to prove that in every solution at least one number is divisible by 3. But I don't see what else can be done to prove that the number of solutions is finite.

My initial question to myself when I saw your post was what limits the solution set to a finite number of elements. My intuition would have been happy with no integer solutions or an infinite number of solutions. And being lazy and entirely innocent of any knowledge of number theory, I decided to work with the simpler problem of a sum of two reciprocals rather than four.

In that simple case, the solution set is built up from integer factors of p. That entails that the solution set has a finite number of elements because the number of integer factors of any finite integer consists of numbers no larger than p.

I admit, however, I am not AT ALL sure my approach works for the sum of four reciprocals of integers. So my approach is nothing more than a suggestion based on finding a reason why the solution set to a simpler problem does not have an infinite number of elements.

One reason to be suspicious of my approach is that p cannot be prime, and in the given problem, p is prime.

Another reason to be suspicious is that my intuition does not let me see a way to extend the result from the sum of k reciprocals to k + 1 reciprocals.

A third reason to be suspicious is that I really do not know any number theory.

 

[Cubist]

I cannot prove it, but my computer believes that the 339 quadruplets listed in the attached files are all possible solutions to that equation. I would be very interested to hear if anyone can either find other solutions or can prove that none exist.

The following python code prints all possible combinations (unless it contains a bug), and it finds 339 which agrees with @blamocur . Obviously many of these could be permuted between x,y,z,t for example {10, 12, 12, 15}, {10, 12, 15, 12}, {10, 15, 12, 12}, etc but this code only prints one permutation (the one with the values in ascending order).

Spoiler: show code

from fractions import Fraction

def subtractFraction(last_denom, num_fract_remaining, sub_from, history_str):
    # The highest possible value for the denominator...
    #
    #  1/highest_denominator + 1/highest_denominator + ... = sub_from
    #
    #    num_fract_remaining / highest_denominator = sub_from
    #
    #    highest_denominator = num_fract_remaining / sub_from
    #
    # Now calculate the highest possible (integer) denominator...
    n = num_fract_remaining * sub_from.denominator // sub_from.numerator
  
    while n >= last_denom:
        fr = Fraction(1, n)
        r = sub_from - fr
        if r < 0: break
        hs = history_str + str(n)
        if num_fract_remaining == 1:
            if r == 0:
                print(hs)
        elif r>0:
            hs = hs + "  "
            subtractFraction(n, num_fract_remaining-1, r, hs)
        n -= 1

subtractFraction(1, 4, Fraction(1,3), "");

 

[Otis]

The following python code prints all possible combinations

What's the largest Integer printed, Cubist?

 

[Cubist]

What's the largest Integer printed, Cubist?

The output is the same as post#7 (but in a different order). You can paste the code into the following site to run it if you don't have Python installed:-
https://www.online-python.com/

Here's some improved code...

Spoiler: show code

You can easily change this for different numbers of "Egyptian fractions" that sum to a given amount by editing the last line.

from fractions import Fraction

def subtractFraction(last_denom, num_fract_remaining, sub_from, history_str):
    if num_fract_remaining == 1:
        if sub_from.numerator == 1:
            print(history_str, sub_from.denominator)
    else:
        # Find the highest possible value for the denominator...
        #
        #  1/highest_denominator + 1/highest_denominator + ... = sub_from
        #
        #    num_fract_remaining / highest_denominator = sub_from
        #
        #    highest_denominator = num_fract_remaining / sub_from
        #
        # Now calculate the highest (integer) denominator...
        n = num_fract_remaining * sub_from.denominator // sub_from.numerator
       
        while n >= last_denom:
            fr = Fraction(1, n)
            r = sub_from - fr
            if r <= 0: break
            hs = history_str + str(n) + "  "
            subtractFraction(n, num_fract_remaining-1, r, hs)
            n -= 1

subtractFraction(1, 4, Fraction(1,3), "");

Top and tail of the output...

12  12  12   12
10  12  12   15
10  10  15   15
10  10  12   20
10  10  10   30
9  12  12   18
...
4  13  160   6240
4  13  159   8268
4  13  158   12324
4  13  157   24492

 

[blamocur]

The following python code prints all possible combinations (unless it contains a bug), and it finds 339 which agrees with @blamocur . Obviously many of these could be permuted between x,y,z,t for example {10, 12, 12, 15}, {10, 12, 15, 12}, {10, 15, 12, 12}, etc but this code only prints one permutation (the one with the values in ascending order).

Spoiler: show code

from fractions import Fraction

def subtractFraction(last_denom, num_fract_remaining, sub_from, history_str):
    # The highest possible value for the denominator...
    #
    #  1/highest_denominator + 1/highest_denominator + ... = sub_from
    #
    #    num_fract_remaining / highest_denominator = sub_from
    #
    #    highest_denominator = num_fract_remaining / sub_from
    #
    # Now calculate the highest possible (integer) denominator...
    n = num_fract_remaining * sub_from.denominator // sub_from.numerator
 
    while n >= last_denom:
        fr = Fraction(1, n)
        r = sub_from - fr
        if r < 0: break
        hs = history_str + str(n)
        if num_fract_remaining == 1:
            if r == 0:
                print(hs)
        elif r>0:
            hs = hs + "  "
            subtractFraction(n, num_fract_remaining-1, r, hs)
        n -= 1

subtractFraction(1, 4, Fraction(1,3), "");

Your Python code works much faster than my optimized C++

 

[Cubist]

Your Python code works much faster than my optimized C++

Then don't try the improved code in post#13 because that's even faster

 

[JeffM]

First, there is an error in my post #8. Let’s get that out of the way initially.

[math]p \text { and } \lambda \text { are positive integers.}[/math]
[math]\sum_{i=1}^{\lambda} \dfrac{1}{\lambda p} = \lambda * \dfrac{1}{\lambda p} = \dfrac{1}{p}.[/math]
Thus, my conclusion that if lambda is 2 and p is prime, there is no pair of reciprocals of positive integers that sum to the reciprocal of p was wrong. There is such a pair. For example

[math]\dfrac{1}{6} + \dfrac{1}{6} = \dfrac{2}{6} = \dfrac{1}{3}.[/math].

Similarly

[math]\dfrac{1}{9} + \dfrac{1}{9} + \dfrac{1}{9} = \dfrac{1}{3} = \sum_{i=1}^4 \dfrac{1}{12}.[/math]
Second, where did my error arise?

I got to [imath]p = j(k - 1)[/imath] and then said that if p was prime, there were no such integers. But that is idiocy. j can be p and (k - 1) can be 1.

But previously I had shown that [imath]p = n - \dfrac{n}{k}.[/imath] And k - 1 = 1 entails that k = 2.

[math]p = n - \dfrac{n}{2} \implies p = \dfrac{n}{2} \implies n = 2p \implies[/math][math]\dfrac{1}{m} + \dfrac{1}{2p} = \dfrac{1}{p} = \dfrac{2}{2p} \implies \dfrac{1}{m} = \dfrac{1}{2p} \implies m = n = \lambda p.[/math]
The error is resolved.

What we can say now.

For any positive integers p and [imath]\lambda[/imath], there exists at least one lambda-tuple of positive integers (not necessarily distinct) whose reciprocals sum to the reciprocal of p. Moreover, if [imath]\lambda[/imath] equals 1 or 2, then the set of such lambda-tuples has a finite number of elements.

All that remains is to show that the finite number of tuples applies no matter what [imath]\lambda[/imath] is.