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Questions and important info in italics, very important ones in bold.

Here we have the system;

$V_{1}+V_{2}\cdots+V_{k}=A$ and $V_{1}^{2}+V_{2}^{2}+\cdots +V_{k}^{2}=B$ where $V_{1}$, $V_{2}$, etc. are distinct, positive integer variables.

According to my previous thread, Jykri Lahtonen assumes that the number of common solutions for the two equations as $k$ increases remains linear in the system, excluding permutations.

But how much, exactly, excluding permutations of values of the variables in the solution across them (simply divide by $k!$)?

And, how do you solve the two equations? Since, even if one of the many solutions are found, I assume the rest can be found easily applying certain equations. For example (which I later found was already stated by Thomas Andrews, IIRC), for any one set of variables $V_{1}$, $V_{2}$, etc. you can observe;

$$V_1^2+V_2^2+\cdots+V_k^2 = \left(\frac{2\left(V_{1}+V_{2}+\cdots+V_{k}\right)}{k}-V_{1}\right)^{2}+\left(\frac{2\left(V_{1}+V_{2}+\cdots+V_{k}\right)}{k}-V_{2}\right)^2+\cdots+\left(\frac{2\left(V_{1}+V_{2}\cdots+V_{k}\right)}{k}-V_{k}\right)^{2}$$

the resulting values will satisfy the system aforementioned iff twice the average of the variables is a whole number.

Assuming I could employ a computer to solve the system of equations, would it be extremely complex as the value for $k$ grows into the thousands, and so on?

Again, I'm completely lost as to what tags describe this topic perfectly. My apologies.

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    If $A$ and $B$ are fixed, then the number of solutions is zero for large values of $k$, since the left sides of your equations are (considerably) bigger than $k$. If $A$ and $B$ are not fixed, then the number of solutions is infinite, even for $k=3$, as we saw the previous go-round.2012-05-21
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    @Gerry Myerson: I go with the assumption that $A$ and $B$ are fixed. BTW by fixed, can I assume that you mean, 'a constant value?' I suppose so.2012-05-21
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    Wait, how can the number of solutions be zero? Under what circumstances will there be only solution? And how do the number of solutions progress as $k$ decreases?2012-05-21
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    Fix values of $A$ and $B$. Now suppose $k$ is bigger than $A$. Well, each $V_i$ is a positive integer, so each $V_i$ is at least 1, so the sum of the $V_i$ is at least $k$, but that's bigger than $A$, so the first equation has no solution. That's how the number of solutions can be zero, indeed, **must** be zero, if $A$ and $B$ are fixed while $k$ is allowed to grow without bound.2012-05-21
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    How can that be? From your own example, for two equations $a+b+c=12$ and $a^{2}+b^{2}+c^{2}=62$, where (obviously, as we have 3 variables) $k=3$, $A=12$ and $B=62$. Assuming **all values for** $a, b, c$ **are distinct**, we have $a=1$, $b=5$, $c=6$ and $a=2$, $b=3$, $c=7$ (the second solution which is calculated from Thomas Andrews' formula: $$V_1^2+V_2^2+\cdots+V_k^2 = \left(\frac{2\left(V_{1}+V_{2}+\cdots+V_{k}\right)}{k}-V_{1}\right)^{2}+\left(\frac{2\left(V_{1}+V_{2}+\cdots+V_{k}\right)}{k}-V_{2}\right)^2+\cdots+\left(\frac{2\left(V_{1}+V_{2}\cdots+V_{k}\right)}{k}-V_{k}\right)^{2}.$$2012-05-21
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    I'd also mentioned in my post that **all values across variables must be distinct.** Keeping that in mind, I don't see how $k$ can exceed $A$ or $B$ unless there is only one variable with a value of 1.2012-05-21
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    You've mucked up the TeX in a comment, making all comments unreadable. I've flagged for moderator's attention.2012-05-22
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    @Gerry: All fixed!2012-05-22
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    You are right, and I am right, and we are saying the same thing, in different ways: you are saying $k$ can't exceed $A$ or $B$, and I am saying if $k$ does exceed $A$ or $B$ then there are no solutions. Two different ways of saying the same thing, right? But your question asks what happens as $k$ increases, so surely we both agree that, as $k$ increases (with $A$ and $B$ held fixed), the number of solutions becomes zero. @Zev, many thanks.2012-05-22
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    Aren't we deviating from the topic? Supposing $A$ and $B$ are fixed, and we first **find out the number of solutions both equations have**, how do you **find the solutions**? And, can all other solutions be constructed from just a handful or one solution as does Thomas Andrews' equation? **Is this computationally intensive?**2012-05-22
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    @Gerry Myerson: I saw that TeX error some hours later. So I couldn't edit.2012-05-22
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    In general, I don't know any way to find the number of solutions other than to find the solutions and count them. In the case $k=3$, my recollection from the other thread is that it had to do with the number of representations as $x^2+xy+y^2$, which in turn has to do with the number of prime divisors that are 1 modulo 3; also, in that case you can generate all the solutions from a few of them. I'll refresh my memory of the earlier discussion and try to write something up when I get a chance.2012-05-23
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    For $k$ going into the billions and not exceeding $A$ or $B$, how computationally intensive could this be, assuming we have some $s$ solutions?2012-05-23

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