Cute problem. Note that $v_i$ can only be one of two values (since the given equation is a quadratic whose coefficients are the same for all $v_i$). You can let the solutions be $a,b$. Let the sum of the $v_i's$ be $S$, the sum of squares of the $v_i's$ be $T$. We get in fact $ab=T/6$, $a+b=T/6$. Summing all the 10 equations, you get $S=10+6T/T=16$. Suppose $m$ of the $v_i's$ equal $a$, the other $n=10-m$ equal $b$. So we have $ma+nb=S=16$ and $ma^2+nb^2=T$. Note that we have $5$ variables $a,b,m,n,T$, and $5$ equations of degree at most $2$, and we can solve by simple elimination/substitution. Actually it turns out really nicely, you get what appears to be an equation of degree $3$ in $a$, but its actually of degree 1, and you find $a=8/5$, $m=10$,$n=0$, so all the $v_i$'s are equal to $8/5$. Note than if you solve the system via elimination/substitution then you have to make the assumption that $v_i$'s are not equal to zero or 1. This is obvious since each $v_i$ is at least 1, and if any $v_i$ equals $1$, then $v_i=0$ (look at the original equation).

Actually $(4, \frac{4}{3}, ..., \frac{4}{3})$ is a solution.

Oops! You are right. Let me see where I made a mistake...

Ah I seem to have made a FATAL mistake. So here is a different approach. Consider just the first equation: $ma+nb=16$. We know $a$, $b$ satisfy the quadratic $x^2=1+6x^2/T$. From this we can deduce $b=a/(a-1)$. We know $a$ is not $1$ so no problem. Thus $ma+(10-m)(a/(a-1))=16$ This is a quadratic in $a$. Solve it: $a=\frac{1}{m}(m+3) \pm \sqrt{m^2-10m+9}$ Since $a$ is real, the discriminant is nonnegative, so $m \le 1$ OR $m \ge 9$. Of course the cases are symmetric (swap $m$ with $n$), so we only need to consider $m=9$ and $m=10$. If $m=10$, all the $v_i$'s are equal, and we easily find $v_i=8/5$. If $m=9$, we get $a=4/3$. So these are the only two solutions.

Denoting $C = \sum_{i=1}^{10} v^2_i$, the $v_i$s are a solution of $\frac{6}{C}x^2-x+1=0$. Therefore, $v_i \in \{a,b\}$. Say there are $u$ $a$s and $v$ $b$s. Vieta and trivial stuff gives $ua+vb=16$, $u+v=10$, $a+b=ab=\frac{C}{6}=\frac{ua^2+vb^2}{6}$. Plug $b=\frac{a}{a-1}$ and $v=10-u$ and solve for $a$. Discriminant gives $u \le 1$ or $u \ge 9$. WLOG, $u \ge 9$. If $u=9$, we have $9a+b=16$, $a+b=ab=\frac{9a^2+b^2}{6}$, so $3a=b$ instantly, giving $(4,\frac{4}{3}, \cdots , \frac{4}{3})$. If $u=10$, we have $a=1+\frac{6}{10} = \frac{8}{5}$, so we have $(\frac{8}{5}, \frac{8}{5}, \cdots , \frac{8}{5})$.

Let $Q = v_1^2 + v_2^2 + \ldots + v_{10}^2$. Rewrite all $10$ equations as\[(v_i - 1)Q = 6v_i^2\]for all ten $i$. Clearly $Q$ is non-negative hence all $v_i > 1$, so we can divide both side to get\[Q = \frac{6v_i^2}{v_i - 1}\]for all ten $i$. Let $S_i = \tfrac{6v_i^2}{v_i - 1}$. Claim: For any $i \neq j$, we have $S_i = S_j$ if and only if $v_i = v_j$ or $v_i = \tfrac{v_j}{v_j - 1}$. Proof: This is actually quite easy. Equate some two $S_i$ and $S_j$ and expand to get $v_iv_j(v_i - v_j) = (v_i - v_j)(v_i + v_j)$ which holds if and only if we have $v_i = v_j$ or $v_iv_j = v_i + v_j \iff v_i = \tfrac{v_j}{v_j - 1}$ for $v_i, v_j > 1$. $\square$ In fact there must be some two $v_i, v_j$ that are equal. Suppose not, and any three $v_i, v_j, v_k$ are all pairwise distinct. However, all $S_n$ are equal to each other hence $v_i = \tfrac{v_j}{v_j - 1} = \tfrac{v_k}{v_k - 1} \implies v_j = v_k$, a contradiction. Therefore, we must have two $v_i = v_j = a$ where $a$ is some real value greater than $1$. Then, for any other $v_k$, we must either have $v_k = a$ or $v_k = \tfrac{a}{a-1}$. Hence, we can partition the set of $\{v_1, v_2, \ldots v_{10}\}$ into two $k$ and $10-k$ size multisets such that one of them consists only of $a$'s and the other consists only of $\tfrac{a}{a-1}$'s. Now plugging everything back in:\[S = ka^2 + (10-k)\left(\frac{a}{a-1}\right)^2 = \frac{6a^2}{a-1} \implies (10-k)r^2 -6r + k = 0\]where we let $r = \tfrac{1}{a-1}$. Since $r$ is real, the discriminant of this quadratic is nonnegative, hence $36 \geq 4k(10-k) \implies k = 0, 1, 9, 10$. Since the $0,10$ and $1, 9$ cases are the same, it only suffices to check for $k = 0$ and $k = 1$. If $k = 0$, then all of them are the same, and thus we get $(v_1, v_2, \ldots v_{10}) = (\tfrac85, \tfrac85, \ldots ,\tfrac85)$. If $k = 1$, then we get that $r = \tfrac13$ hence $a = 4$ and $\tfrac{a}{a-1} = \tfrac43$. Thus, in this case $(v_1, v_2, \ldots v_{10})=(\tfrac43, \tfrac43, \ldots ,\tfrac43, 4)$ or one of its permutations. Now we are done. $\blacksquare$ Remark: Overcomplicated eh? But I think it's well explained.

AlcumusGuy wrote: Consider the following system of $10$ equations in $10$ real variables $v_1, \ldots, v_{10}$: \[v_i = 1 + \frac{6v_i^2}{v_1^2 + v_2^2 + \cdots + v_{10}^2} \qquad (i = 1, \ldots, 10).\]Find all $10$-tuples $(v_1, v_2, \ldots , v_{10})$ that are solutions of this system. AYYYY u were in my geo, or algebra B cLASS ONCE

Summing these equations together gives us: $$ v_1+v_2+ \ldots+v_{10} = 16 $$The crux of the problem is the following claim. Claim: The set of the values of $v_1, v_2, \ldots, v_{10}$ consists of at most $2$ different numbers. Proof: Consider arbitrary indices $i,j$. We have: $$ v_i = 1 + \frac{6v_i^2}{v_1^2 + v_2^2 + \cdots + v_{10}^2} \qquad \text{and} \qquad v_j = 1 + \frac{6v_i^2}{v_1^2 + v_2^2 + \cdots + v_{10}^2} $$Subtracting these $2$ equations yields to: $$ S(v_i - v_j) =6(v_i-v_j)(v_i+v_j) $$where $S=v_1^2+v_2^2+ \ldots + v_{10}^2$. This means that either $v_i = v_j$ or $S=6(v_i+v_j)$. If we fix $v_1$, then we get the values that are equal to $v_1$ and some that are not. We will prove that those, which are not equal to $v_1$, are equal between themselves. Suppose there are $2$ indices $i,j$ such that $v_1 \neq v_i$ and $v_i \neq v_j$, then $S=6(v_1+v_i)=6(v_1+v_j) \implies v_i = v_j$. This proves the claim. This means that there are $x$ numbers, which attains the value $a$ and $10-x$ numbers which attains the value $b$. From before established results we get the following equations. $$ xa+(10-x)b = 16 \qquad \text{and} \qquad xa^2+(10-x)b^2 =6(a+b) $$First equation can be rewritten as follows: $$ x(a-b) = 16 -10b $$This means that: \begin{align*} x(a-b)(a+b) =6(a+b)-10b^2 \\ (16-10b)(a+b) = 6(a+b)-10b^2 \\ (8-5b)(a+b) = 3(a+b)-5b^2 \\ 8(a+b) -5ab = 3(a+b) \\ a+b = ab \\ b = \frac{a}{a-1} \end{align*}Plugging this back into the first equation yields to: \begin{align*} ax+(10-x)\frac{a}{a-1} = 16 \\ a(a-1)x + (10-x)a = 16(a-1) \\ a^2x -ax +10a - xa =16a - 16 \\ xa^2 -(2x+6)a+16 =0 \\ \end{align*}Treating this as a quadratic equation with respect to $a$, gives us tht its discriminant must be nonnegative; therefore: $$ (2x+6)^2 - 64x \ge0 \implies 4(x-1)(x-9) \ge 0 $$Thus $x=0;1;0;10$, which gives us solutions $v_1, v_2, \ldots v_{10}) = (\frac{8}{5}, \frac{8}{5}, \ldots ,\frac{8}{5})$ and $(v_1, v_2, \ldots v_{10})=(\frac{4}{3}, \frac{4}{3}, \ldots ,\frac{4}{3}, 4)$ and it permutations.

If there exist a triple $(i,j,k)$ of distinct indexes such that $v_i\ne v_j\ne v_k$, we get: $$v_k-v_j=\frac{6}{v_1^2+\dots+v_{10}^2}(v_k-v_j)(v_k+v_j)\implies v_k+v_j=\frac{v_1^2+\dots+v_{10}^2}{6},$$and $$v_j-v_i=\frac{6}{v_1^2+\dots+v_{10}^2}(v_j-v_i)(v_j+v_i)\implies v_j+v_i=\frac{v_1^2+\dots+v_{10}^2}{6}$$which implies that $v_k=v_i$. Therefore, we have at most two values for the $v_i$'s, say $a\leq b$. By summing up all the equations we get $ka+(10-k)b=16\implies k(a-b)=16-10b$, where $k$ is the amount of $v_i=a$. Suppose $a\ne b$. By subtracting $a-b$ we get: $$a+b=\frac{ka^2+(10-k)b^2}{6}=\frac{16(a+b)-10ab}{6}\implies a+b=ab\implies b=\frac{a}{a-1} (i).$$From $a=1+\frac{6a^2}{ka^2+(10-k)b^2}$ and $(i)$ we get: $$a-1=\frac{6a^2(a-1)^2}{k(a^4-2a^3)+10a^2}=\frac{6(a-1)^2}{ka^2-2ka+10}$$$$\implies(a-1)\left(1-\frac{6(a-1)}{ka^2-2ka+10}\right)=0,$$and since $a=1$ is obviously a contradiction $$a^2k-a(2k+6)+16=0\implies a=\frac{2k+6\pm\sqrt{(2k+6)^2-64k}}{2k}$$$$\implies(2k+6)^2-64k\leq0\implies(k-9)(k-1)\leq0\implies k=1\text{ or }k=9.$$If $k=1$, we get $a=4$ which implies a contradiction from $ka+(10-k)b=16$ and $b\geq a$. If $k=9$, we get $a=\frac{4}{3}$ and $b=4$, which is a solution. If $a=b$ we get $a=b=\frac{8}{5}$, which is also a solution. From this we conclude that all solutions are $\left(\frac{8}{5},\frac{8}{5},\dots,\frac{8}{5}\right)$ and $\left(\frac{4}{3},4,4,\dots,4\right)$ and its permutations. $\blacksquare$

We can rearrange this to get $A=v_1^2+\dots+v_{10}^2=\frac{6v_i^2}{v_i-1}$ for all $i$. Then, $A=\frac{A(v_1-1)}{6}+\dots+\frac{A(v_{10}-1)}{6}$, so $6=v_1+\dots+v_{10}-10$ so $v_1+\dots+v_{10}=16$. By Jensen's, $A\ge 25.6$. This also means that $v_i=\frac{A\pm \sqrt{A^2-24A}}{12}$ for all $v_i$. Let $5-x$ of the $v_i$ be $\frac{A- \sqrt{A^2-24A}}{12}$ and $5+x$ of the $v_i$ be $\frac{A+ \sqrt{A^2-24A}}{12}$. Then, we must have that $\frac{5A}{6}+\frac{x\sqrt{A^2-24A}}{6}=16$ or $5A-96=x\sqrt{A^2-24A}$. Since $5A-96\ge 5\cdot25.6-96>0$, $x>0$. If $x=5$, we get the solution $A=25.6$, corresponding to the $10$-tuple $(1.6, \dots, 1.6)$. Otherwise, since $\sqrt{A^2-24A} < A-12$, $5A-96<xA-12x$ so $A<\frac{96-12x}{5-x}$. Since $A\ge 25.6$, $x=3$ or $x=4$. We simplify $5A-96=x\sqrt{A^2-24A}$ to $25A^2-960A+96^2=x^2(A^2-24A)$ or $(25-x^2)A^2+(24x^2-960)A+96^2=0$. When $x=3$, the discriminant is $744^2-64\cdot96^2=744^2-768^2<0$. When $x=4$, our quadratic becomes $9A^2-576A+96^2=0$ so $A^2-64A+32^2=0$ so $A=32$, corresponding to $10$-tuple $(4, \frac43, \dots, \frac43)$ and permutations. Therefore, the solutions are $(1.6, \dots, 1.6)$ and $(4, \frac43, \dots, \frac43)$ and permutations.

Define \[S = v_1^2 + v_2^2 + \cdots + v_{10}^2 \implies v_i = 1+\frac{6v_i^2}{S}\]Solving the quadratic gives \[v_i = \frac{S\pm\sqrt{S^2-24S}}{12} \implies v_i^2 = \frac{S^2\pm 2S\sqrt{S^2-24S}+S^2-24s}{144}\]Now if there are $p$ number of $v_i$ that are $+$ and $m$ that are $-$, then the sum of $v_i^2$ is \[S = \sum_{i = 1}^{10} v_i^2 = \frac{20S^2-240S+2kS\sqrt{S^2-24S}}{144}\]\[\implies k^2(S^2-24S) = (192-10S)^2\]where $k = |p-m|$. Taking the discriminant gives \[k^2(k^2-64) \ge 0 \implies k^2 \ge 64 \text{ or } k^2 \le 0 \implies k = 0,8,10\]Solving these cases for $k$ gives the answer \[\boxed{v_i = 4 \text{ for all $i$ except one of them is } v_j = \frac{4}{3} \text{ or } v_i = \frac{8}{5} \text{ for all $i$}}\]

We claim that all solutions are $\left(\frac{8}{5}, \frac{8}{5},\ldots, \frac{8}{5}\right)$ and permutations of $\left(4, \frac{4}{3}, \ldots, \frac{4}{3}\right)$. First, summing over all $i$ gives us \[\sum_{i=1}^{10} v_i = 16.\]Now, note that $v_i>1$ for all $i$, then we have for some $i,j$, \[\frac{v_i}{v_j}=\frac{t+6v_i^2}{t+6v_j^2}\]where $t=\sum_{i=1}^{10} v_i^2$, which then gives us \[\frac{t}{v_i} + 6v_i = \frac{t}{v_j} + 6v_j \iff t\cdot \frac{v_j-v_i}{v_iv_j}=6(v_j-v_i).\]Thus, either $v_i=v_j$, or $t=6v_iv_j$. In the case where $v_i=v_j$ for all $i,j$, we have that there is only one $10$-tuple, which must be $\left(\frac{8}{5}, \frac{8}{5},\ldots, \frac{8}{5}\right)$. In the case that $v_i\neq v_j$ for some $i,j$, it is first clear that we cannot have $v_i>v_j>v_k$ for some $i,j,k$, because otherwise we would have $t = 6v_iv_j > 6v_jv_k = t$, which is a contradiction. So, we can only possibly have $v_i=a$ for some $i$s and $v_i=b$ for other $i$s. WLOG the number of $a$s is at least the number of $b$s. We have the equations \[a=1+\frac{a}{b}, b = 1+\frac{b}{a} \implies ab = a+b.\]Now, $6ab=ka^2+(10-k)b^2$, and since $a^2+b^2\geq 2ab$, we cannot have $3$ of either $a$ or $b$, so either $9a+b=16$ or $8a+2b=16$. Substituting $a=\frac{b}{b-1}$ into these equations, we get $(a,b) = \left(\frac{4}{3}, 4\right)$ and no solution for the second equation. Then, we can see that all $10$ permutations of the $10$-tuple $\left(4, \frac{4}{3}, \ldots, \frac{4}{3}\right)$ indeed works, so we are done. Note: I thought this was a very nice problem, and the motivation of considering $\frac{v_i}{v_j}$ comes from the fact that the RHS would be $\frac{v_i^2}{v_j^2}$ if the $1$s weren't there. Intuitively, there aren't many ways the RHS which contains $\frac{v_i^2}{v_j^2}$ can become $\frac{v_i}{v_j}$.

By summing it is obvious that $\sum v_i=16$. Let \[ v_i=m_i+1 \]Now we have $\sum m_i=6$. We have \[ \frac{m_i}{(m_i+1)^2}=\frac{6}{(m_1+1)^2+(m_2+1)^2+\dots +(m_{10}+1)^2} \]Since the rhs is equal for all $i\in \{ 1,2,\dots ,10\}$ we have \[ \frac{m_i}{(m_i+1)^2}=\frac{m_j}{(m_j+1)^2} \]And after doing some simple algebra we end up with \[ (m_i-m_j)(m_im_j-1)=0 \]Hence for any pair of $m_i$ and $m_j$ we have that they are either equal or their multiplication is $1$. We know that the multiplication of numbers is only equal to $1$ if one of the numbers is larger than $1$ and the other is smaller then $1$. OR they are both equal to $1$. So, since the numbers are either equal or when multiplied equal to $1$, we know that if they are both smaller than $1$, then they are equal. Alternatively, if they are both larger then $1$ then are also equal. If a number is larger then $1$, let it have the value $a$, if the number is smaller then $1$ let it have the value $b$. Assume we have $x$ numbers with value $a$, $y$ numbers with value $b$ and $z$ numbers with value $1$. Of course $x+y+z=10$ and $xa+yb+z=6$. The cool thing that we have now is that $x,y,z$ are positive integers, moreover, we can say that $x<6$, so we can kill this with case work! Also, note that when we combine $x+y+z=10$ and $xa+yb+z=6$ along with $ab=1$, we get a quadratic whoose discriminant must be $\geq 0$. In algebra, this is: \[ (6-z)^2-4xy\geq 0 \]\begin{itemize} \ii Let $x=5$. Now the only choice for $z$ is $0$ which does not work \ii Let $x=4$. We proceed by similar reasoning, but this time we divide the case into smaller cases when $z=0$ and $z=1$. None of them work. \ii Let $x=3$. No case works. \ii Let $x=2$. Again, nothing. \ii Let $x=1$. Here, if we let $z=0$ and $y=9$ we actually get something that works! (it is the only thing that works in this case). We get $a=3$ and $b=\frac{1}{3}$, so in turn we get $v_i=4$ for a single $i$ and $v_j=\frac{4}{3}$ for all $j\neq i$ However, upon putting it into the initial equation, we find out that the solution found above indeed is a correct solution. \ii Let $x=0$. We now have that. Now, since $z\leq 5$ (as can be seen in $xa+yb+z=6$). We have to try the cases \[ (y,z)=(5,5),(6,4),(7,3),(8,2),(9,1),(10,0) \]And after trying everything out we see that $(y,z)=(10,0)$ is the only one to produce a solution. Here we get $b=\frac{3}{5}$ which then gives us that $v_i=\frac{8}{5}$ for all $i$. \end{itemize} Thus the only possible solutions are $(4, \frac{4}{3}, ..., \frac{4}{3})$ and $(\frac{5}{8},\frac{5}{8},\dots ,\frac{5}{8})$