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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. – Gerry Myerson May 21 '12 at 11:15
    
@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. – Mach9 May 21 '12 at 12:01
    
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? – Mach9 May 21 '12 at 12:08
    
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. – Gerry Myerson May 21 '12 at 13:06
    
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: $$V_1^2+V_2^2+\cdots+V_k^2 = \left(\frac{2\left(V_{1}+V_{2}+\cdots+V_{k}\right)}{k}-V_{1}\right)^{2}+\left(\f‌​rac{2\left(V_{1}+V_{2}+\cdots+V_{k}\right)}{k}-V_{2}\right)^2+\cdots+\left(\frac{‌​2\left(V_{1}+V_{2}\cdots+V_{k}\right)}{k}-V_{k}\right)^{2}.$$ – Mach9 May 21 '12 at 13:15

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