As many users have not been exposed to complex numbers, I will post the solutions to the problems assuming that all variables used are real. However, please feel free to solve these problems over the complex numbers and post your solutions!

**Problem 1.** Solve

$\left\{\begin{aligned}y^2&=x^3-3x^2+2x&\quad(1)\\x^2&=y^3-3y^2+2y&\quad(2)\end{aligned}\right.$

**Solution.** $(1)-(2)$:

$-(x^2-y^2)=(x^3-y^3)-(3x^2-3y^2)+(2x-2y)$

$-(x+y)(x-y)=(x-y)(x^2+xy+y^2)-(3x+3y)(x-y)+2(x-y)$

$(x-y)(x^2+xy+y^2-3x-3y+2+x+y)=0$

$(x-y)[x^2+(y-2)x+(y^2-2y+2)]=0$

**Case 1.** $x-y=0\implies x=y$. Substituting $x=y$ into $(2)$, we have

$y^2=y^3-3y^2+2y$

$y(y^2-4y+2)=0$

**Case 1a.** $y=0$. Since $x=y$, $\boxed{x=y=0}$.

**Case 1b.** $y^2-4y+2=0$.

$y = \frac{-(-4)\pm\sqrt{(-4)^2-4(1)(2)}}{2(1)}$

$=2\pm\sqrt{2}$

Since $x=y$, $\boxed{x=y=2\pm\sqrt{2}}$.

**Case 2.** $x^2+(y-2)x+(y^2-2y+2)]=0$.

$\text{Discriminant}\,= (y-2)^2-4(1)(y^2-2y+2)$

$=y^2-4y+4-4y^2+8y-8$

$=-3y^2+4y-4$

$=-3\left(y^2-\frac{4}{3}y\right)-4$

$=-3\left[y^2-\frac{4}{3}y+\left(\frac{2}{3}\right)^2-\left(\frac{2}{3}\right)^2\right]-4$

$=-3\left[\left(y-\frac{2}{3}\right)^2-\frac{4}{9}\right]-4$

$=-3\left(y-\frac{2}{3}\right)^2+\frac{4}{3}-4$

$=-3\left(y-\frac{2}{3}\right)^2-\frac{8}{3}<0$

Since we are solving this problem over the reals, Case 2 is rejected.

Hence, $\boxed{x=y=0}$ and $\boxed{x=y=2\pm\sqrt{2}}$.

**Problem 2.** Solve

$\left\{\begin{aligned}5x^2+4y^2-10x+5y&=0&\quad(3)\\4x^2+5y^2+5x-10y&=0&\quad(4)\end{aligned}\right.$

**Problem 3.** Solve

$\left\{\begin{aligned}x+y-z&=4&\quad(5)\\x^2+y^2-z^2&=12&\quad(6)\\x^3+y^3-z^3&=34&\quad(7)\end{aligned}\right.$

**Problem 4.** Solve

$\left\{\begin{aligned}xy+yz+zx&=1&\quad(8)\\yz+zt+ty&=1&\quad(9)\\zt+tx+xz&=1&\quad(10)\\tx+xy+yt&=1&\quad(11)\end{aligned}\right.$

**Problem 5.** Solve

$\left\{\begin{aligned}(x+y)(x+z)&=x&\quad(12)\\(y+z)(y+x)&=2y&\quad(13)\\(z+x)(z+y)&=3z&\quad(14)\end{aligned}\right.$

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## Comments

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TopNewestProblem 1.Solve$\left\{\begin{aligned}y^2&=x^3-3x^2+2x&\quad(1)\\x^2&=y^3-3y^2+2y&\quad(2)\end{aligned}\right.$

Solution.$(1)-(2)$:$-(x^2-y^2)=(x^3-y^3)-(3x^2-3y^2)+(2x-2y)$

$-(x+y)(x-y)=(x-y)(x^2+xy+y^2)-(3x+3y)(x-y)+2(x-y)$

$(x-y)(x^2+xy+y^2-3x-3y+2+x+y)=0$

$(x-y)[x^2+(y-2)x+(y^2-2y+2)]=0$

Case 1.$x-y=0\implies x=y$. Substituting $x=y$ into $(2)$, we have$y^2=y^3-3y^2+2y$

$y(y^2-4y+2)=0$

Case 1a.$y=0$. Since $x=y$, $\boxed{x=y=0}$.Case 1b.$y^2-4y+2=0$.$y = \frac{-(-4)\pm\sqrt{(-4)^2-4(1)(2)}}{2(1)}$

$=2\pm\sqrt{2}$

Since $x=y$, $\boxed{x=y=2\pm\sqrt{2}}$.

Case 2.$x^2+(y-2)x+(y^2-2y+2)]=0$.$\text{Discriminant}\,= (y-2)^2-4(1)(y^2-2y+2)$

$=y^2-4y+4-4y^2+8y-8$

$=-3y^2+4y-4$

$=-3\left(y^2-\frac{4}{3}y\right)-4$

$=-3\left[y^2-\frac{4}{3}y+\left(\frac{2}{3}\right)^2-\left(\frac{2}{3}\right)^2\right]-4$

$=-3\left[\left(y-\frac{2}{3}\right)^2-\frac{4}{9}\right]-4$

$=-3\left(y-\frac{2}{3}\right)^2+\frac{4}{3}-4$

$=-3\left(y-\frac{2}{3}\right)^2-\frac{8}{3}<0$

Since we are solving this problem over the reals, Case 2 is rejected.

Hence, $\boxed{x=y=0}$ and $\boxed{x=y=2\pm\sqrt{2}}$.

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When solving system of equations of the form $x = f(y) , y = f(x)$, it is often very tempting to conclude that the only solution occurs when $x = y$.

What we actually have, is $f(f(x)) = x$, and so we are interested in the cases where $f(x) = f^{-1} (x)$. If you can accurately plot this graph, you can then find all such solutions to this equation.

Otherwise, take the approach as Victor did, where you subtract the 2 equations and factor out $(x - y)$. If you believe that the only solutions occur when $x = y$, then you just need to show that the dividend has no (real) roots.

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Do you think I should continue to post the solutions as comments? Depending on the number of upvotes of each solution, the solutions will not be arranged in chronological order.

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I think that posting the solution below the problem makes it disruptive to seeing the entire list. I would prefer if you simply posted the solution as comments, or at the end of the note.

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This is a nice set of problems to work on. The tricky part is justifying that one has indeed found all of the solutions. Try solving these problems over the reals and over the complex numbers, and see how your solution differs. The first problem is a great example!

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Subtracting the equations in $1$,we get a factorization.One of the factors yields a solution,and looking at the discriminant of the other one yields no solutions(real).

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Indeed, justifying that one has indeed found all of the solutions is the tricky part. As many other users have not been exposed to complex numbers, I will post the solutions to the problems assuming that all variables used are real. However, other users may feel free to solve these problems over the complex numbers and post their own solutions.

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Thanks! Often, the real case gives insight into the approach.

And of course, we can then ask, what are the solutions over the quarternions? Over the matrices? It never ends :)

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FIRST ONE IS X=Y fourth one is x=y=z=t

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Problem 1: y^2=x(x-1)(x-2) and x^2=y(y-1)(y-2). x=y=0. (if there is only one solution) Problem 2: x=y=0. (if that is the only solution) Problem 3: (x,y,z)-->(2,3,1) or (3,2,1) Problem 4: t=x=y=z=sqrt(1/3) Problem 5: x=y=z=0 I assume there is only 1 solution for all. very dangerous. Anyway, I'm actually 12, and i'm going RI too! Hope to see you soon, Victor and Srivathsan Veeramani.

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Good luck aloysius to get into RI quite surprised you knowthis at a young age I only learnt this recently but I feel that these sums will come in smo round 1 which does not requre methods and sometimes it is quite obvious like qs 1 that x=y and I am too lazy to write the method.anyway you have solved it correctly and congrats.Good luck and have fun in RI.

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There are actually several solution sets to question 1. If we were restricted to the real, then there are 3 solutions with $x = y$. If we include the complex numbers, then there are 6 solutions

and$x \neq y$.With these questions, you should think beyond finding the obvious answer, and trying to ensure that you could find all possible solutions.

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(3).

x+y=4+z $x^{2}+y^{2}=12+z^{2}$

After this we can find xy, and hence $x^{3}+y^{3}$ in terms of z. Comparing with the third equation will give a cubic equation.

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