# Number of solutions (excluding permutations of variables' values) and solving in distinct positive integers the following system of equations

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