(most elegant) solution of equation

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ednoc

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What is the most elegant (simple or natural) solution of the equation

a+8b+27c+64d+125e=0
 
The most important question here is, What does "solution" mean here?

If there were one variable, it would mean to find the value(s) of that variable for which the equation is true. But with five variables, there are infinitely many sets of values that would make it true. One of them is that all the variables are zero. But it's only one of many (even if you want the variables all to be integers).

So, what are you really asking for? What is the context of your question? And what have you done to try to solve it (elegantly or not)?
 
ednoc said:
What is the most elegant (simple or natural) solution of the equation
\(\displaystyle a+8b+27c+64d+125e=0\)
Click to expand...
How about a minimal solution: \(\displaystyle a=b=c=d=1~\&~e=\frac{-4}{5}\)
 
MarkFL said:
I was waiting for some clarification from the OP before possibly moving the thread.
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I put it to you, being in the calculus forum will encourage more serious consideration of the issues that this question raises.
 
pka said:
Does that matter? If so, in context does it matter?
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We have a linear equation in five unknowns. The question asked about that equation is, to be polite, vague. We do not even know what number system is involved. The question was posted under calculus. Maybe if we knew why it was plunked down in this forum, we would have some clue on what the question is really asking.

You gave an answer in Q. Why Q? In what sense is your answer supposed to be "minimal"? What is the argument that your answer is a minimum relative to a = 0, b = 0, c = 0, d = 0, and e = 0? Whether or not my answer is minimal in some sense, my answer in Z does seem relatively simple, which was the purported objective of the original post.
 
Several points I should have clarified in my first post:
  1. First, I forgot to exclude the trivial solution a=b=c=d=e=0.
  2. Second, only integers solutions are sought (i.e., it should be viewed as a diophantine equation).
  3. When I said "elegant" --or "natural"--, I was thinking in a simple solution that could be generalized to other analogous problems. (Up to now I cannot be more concrete about the qualifier "analogous")
Thank your for your answers
 
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Given the above clarification, I amend my proposed solution: \(\displaystyle a=26,~b=c=d=1~\&~e=-1\)
Note that there are no solutions in the positive integers.
 
I think the question is still quite unclear. Does "elegant, simple, or natural" refer to the solution itself, or to the method of solution (as I would normally take it, and which at least has some meaning)? What specific criterion would you use to judge one solution as better than another? Do you want the smallest possible numbers, for example?

And, of course, until you say what you mean by "analogous problems", you have not said anything about your goal. Deliberately withholding part of the statement of a problem is not good mathematics.
 
ednoc said:
Several points I should have clarified in my first post:
  1. First, I forgot to exclude the trivial solution a=b=c=d=e=0.
  2. Second, only integers solutions are sought (i.e., it should be viewed as a diophantine equation).
  3. When I said "elegant" --or "natural"--, I was thinking in a simple solution that could be generalized to other analogous problems. (Up to now I cannot be more concrete about the qualifier "analogous")
Thank your for your answers
Click to expand...
This at least does something to narrow what your question means (although I continue to wonder if calculus has any relevance). I point out that there may be no answer, let alone a "simple" one, and that the concept of generalizing is not relevant until a class of problems is clearly specified.

At this point, I am guessing that the problem is something like

[MATH]\text {Given } i,\ n \in \mathbb Z^+,\ 1 \le i \le n,\ x_i \in \mathbb Z,\ x_i \ne 0, \text { and}[/MATH]
[MATH]\left ( \sum_{i=1}^n i^3x_i \right ) = 0, \text { find a solution.}[/MATH]
It should be obvious that there is no solution if n = 1. So maybe we should specify in the problem that n > 1.

If n = 2, we are looking for

[MATH]x_1 \text { and } x_2 \text { such that } x_1 + 8x_2 = 0, x_1,\ x_2 \in \mathbb Z, x_1 \ne 0, \text { and } x_2 \ne 0.[/MATH]
There are an infinite number of answers, all in the form:

[MATH]\dfrac{x_1}{x_2} = -\ 8.[/MATH]
If n = 3, there are again an infinite number of answers, at least some of which are in the form:

[MATH]x_1 = x_2 \text { and } \dfrac{x_1}{x_3} = -\ 3. [/MATH]
At this point I can "see" that an answer to the original question as modified is

[MATH]x_1 = x_2 = x_3 = x_4 = 20 \text { and } x_5 = -\ 16 \ \because[/MATH]
[MATH]20 + 8 * 20 + 27 * 20 + 64 * 20 - 125 * 16 = 100 * 20 - 2000 = 0.[/MATH]
 
Dr.Peterson said:
I think the question is still quite unclear. Does "elegant, simple, or natural" refer to the solution itself, or to the method of solution (as I would normally take it, and which at least has some meaning)? What specific criterion would you use to judge one solution as better than another? Do you want the smallest possible numbers, for example?

And, of course, until you say what you mean by "analogous problems", you have not said anything about your goal. Deliberately withholding part of the statement of a problem is not good mathematics.
Click to expand...

Thank you all for your answers.

When I said "analogous problems" I had in mind the following:

a+8b+27c+64d+125e=0
a+8b+27c+64d+125e+216f=0
a+8b+27c+64d+125e+216f+343g=0
... and so on
 
ednoc said:
Thank you all for your answers.

When I said "analogous problems" I had in mind the following:

a+8b+27c+64d+125e=0
a+8b+27c+64d+125e+216f=0
a+8b+27c+64d+125e+216f+343g=0
... and so on
Click to expand...
Which is specifically the class that I addressed.
 
[MATH]\text {Given } \left ( \sum_{i=1}^n i^3x_1 \right ) = 0, x_i \ne 0, \text { and } n \text { is an integer } > 1,[/MATH]
[MATH]\text {find a solution.}[/MATH]
One solution is to have the first n x_i all equal the same integer u.

[MATH]\therefore 0 = \left ( \sum_{i=1}^n i^3x_i \right ) = u * \left ( \sum{i=1}^{n-1} i^3 \right ) + n^3x_n = \dfrac{u(n - 1)^2n^2}{4} + n^3x_n \implies[/MATH]
[MATH]-\ nx_n = \dfrac{u(n - 1)^2}{4} \implies \dfrac{u}{x_n} = - \dfrac{4n}{(n - 1)^2}.[/MATH]
So if n = 2, the smallest possible positive u is 8 with x_2 = - 1 as previously shown.

If n = 3, the smallest possible positive u is 3 with x_3 = - 1 as previously shown.

[MATH]n = 5 \implies \dfrac{u}{x_5} = -\ \dfrac{4 * 5}{4^2} \implies u * \dfrac{4}{5} = -\ x_5.[/MATH]
So u must have 5 as a factor. In which case u = 5 is the smallest positive integer possible, which makes x_5 = - 4.

[MATH]5 + 8 * 5 + 27 * 5 + 64 * 5 - 4 * 125 = 5(1 + 8 + 27 + 64) - 500 = 5 * 100 - 500 = 0.[/MATH]
Of course, these are not unique answers.
 
ednoc said:
a+8b+27c+64d+125e=0
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Just a "looper" solve...
If all 5 variables are restricted to integers -9 to 9, then there are 1047 solutions...
....n.. : a....b....c....d.... e
0001: -9, -9, -4, +1, +1
0002: -9, -9, +3, 0, 0
0003: -9, -8, -7, +8, -2
........
1045: +9, +8, +7, -8, +2
1046: +9, +9, -3, 0, 0
1047: +9, +9, +4, -1, -1
 
Leave it to you to find computationally all the solutions to the original problem. But I think at the end of the day the OP wanted a generic solution to a family of problems. The whole thing would have been more interesting had the OP put more constraints on the problem.
 
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