I have to find all triplets $(x,y,z)$ that satisfy:
$$x^{2012} + y^{2012} + z^{2012} = 3\\x^{2013} + y^{2013} + z^{2013} = 3\\x^{2014} + y^{2014} + z^{2014} = 3$$
I've found the trivial solution $(1,1,1)$ but I don't know how to start looking for more... Does this system have an infinite amount of solutions?
On
While the method already posted is nice, can't resist this hint - from the first and last equation, by the inequality of power means $$\sqrt[2014]\frac{x^{2014}+y^{2014}+z^{2014}}3 \geqslant \sqrt[2012]\frac{x^{2012}+y^{2012}+z^{2012}}3$$
with equality iff $|x|=|y|=|z|=1$. Now using the second equation, the unique answer is obvious.
P.S. This obviously generalises to several variables - all you need are three equations, two of which with even exponents.
Let $e_1$, $e_2$ and $e_3$ be LHS of the the first, second and third equations respectivelly, then:
$$e_3 -2e_2 + e_1 = \\ x^{2012}(x^2 - 2x + 1) + y^{2012}(y^2-2y+1) + z^{2012}(z^2-2z + 1) = 0 \Longrightarrow \\\Longrightarrow x^{2012}(x-1)^2 + y^{2012}(y-1)^2 + z^{2012}(z-1)^2 = 0 $$
Since all the sumands are product of even powers, they cannot be negative, so they all are $0$. That means that each unknown is either $0$ or $1$. But only if they all are $1$ the original equations hold.
You can generalize this to $n$ equations of the same kind and $n$ unknowns:
Then we wan define $e_j$ as the $LHS$ of the $j$-th equation. Then:
$$\sum_{k=1}^n \binom{n}{k}(-1)^{k+1}e_k = \sum_{k=1}^{n}x_k^a(x_k-1)^{n-1} = 0$$
So same as before, each $x_k$ is either $1$ or $0$, but only the $n$-tuple $(1,1,\cdots,1)$ satisfies the equation.
If all the RHS are $m < n$ with $m$ a positive integer, then the solutions are the $n$-tuples with $m$ ones and $n-m$ zeros permutated.