# 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 Trivial solution: (1,1,1)

Find all triplets $\left(x,y,z\right)$ that satisfy:
${x}^{2012}+{y}^{2012}+{z}^{2012}=3\phantom{\rule{0ex}{0ex}}{x}^{2013}+{y}^{2013}+{z}^{2013}=3\phantom{\rule{0ex}{0ex}}{x}^{2014}+{y}^{2014}+{z}^{2014}=3$
Trivial solution: $\left(1,1,1\right)$
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Amiya Watkins
Let ${e}_{1}$, ${e}_{2}$ and ${e}_{3}$ be LHS of the the first, second and third equations respectivelly, then:
${e}_{3}-2{e}_{2}+{e}_{1}=\phantom{\rule{0ex}{0ex}}{x}^{2012}\left({x}^{2}-2x+1\right)+{y}^{2012}\left({y}^{2}-2y+1\right)+{z}^{2012}\left({z}^{2}-2z+1\right)=0⟹\phantom{\rule{0ex}{0ex}}⟹{x}^{2012}\left(x-1{\right)}^{2}+{y}^{2012}\left(y-1{\right)}^{2}+{z}^{2012}\left(z-1{\right)}^{2}=0$
Since all the sumands are product of even powers, they cannot be negative, so they all are $1$. That means that each unknown is either $0$ or $1$. But only if they all are $1$ the original equations hold.
Generalize this to $n$ equations of the same kind and $n$ unknowns:
$\left\{\begin{array}{rcllllllllll}& {x}_{1}^{a}& +& {x}_{2}^{a}& +& {x}_{3}^{a}& +& \dots & +& {x}_{n}^{a}& =& n\\ & {x}_{1}^{a+1}& +& {x}_{2}^{a+1}& +& {x}_{3}^{a+1}& +& \dots & +& {x}_{n}^{a+1}& =& n\\ & {x}_{1}^{a+2}& +& {x}_{2}^{a+2}& +& {x}_{3}^{a+2}& +& \dots & +& {x}_{n}^{a+2}& =& n\\ & ⋮& & ⋮& & ⋮& & \ddots & & ⋮& & ⋮\\ & {x}_{1}^{a+n-1}& +& {x}_{2}^{a+n-1}& +& {x}_{3}^{a+n-1}& +& \dots & +& {x}_{n}^{a+n-1}& =& n\end{array}$
With $a$ even and $n$ odd.
Then we wan define ${e}_{j}$ as the $LHS$ of the $j$-th equation. Then:
$\sum _{k=1}^{n}\left(\genfrac{}{}{0}{}{n}{k}\right)\left(-1{\right)}^{k+1}{e}_{k}=\sum _{k=1}^{n}{x}_{k}^{a}\left({x}_{k}-1{\right)}^{n-1}=0$
So same as before, each ${x}_{k}$ is either $1$ or $0$, but only the $n$-tuple $\left(1,1,\cdots ,1\right)$ satisfies the equation.
If all the RHS are $m with $m$ a positive integer, then the solutions are the $n$-tuples with $m$ ones and $n-m$ zeros permutated.
###### Did you like this example?
Ilnaus5
While the method already posted is nice, can't resist this hint - from the first and last equation, by the inequality of power means
$\sqrt[2014]{\frac{{x}^{2014}+{y}^{2014}+{z}^{2014}}{3}}⩾\sqrt[2012]{\frac{{x}^{2012}+{y}^{2012}+{z}^{2012}}{3}}$
with equality iff $|x|=|y|=|z|=1$. Now using the second equation, the unique answer is obvious.
###### Did you like this example?

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee