Rewrite an expression as a sum of even powers - proving injectivity without calculus I teach algebra to undergraduates - nothing too fancy. But we recently covered injective functions, and I was a little disappointed that I couldn't give rigorous justifications why certain functions were injective. For the handful of functions that they are expected to know, the only justifications I've seen that these functions are indeed injective proceed by showing that they are strictly increasing or decreasing on certain intervals by means of calculus - much more machinery than is covered in this class. However, there was one function which I - excitedly - got to prove to them was injective. Namely $f(x)=x^3$. For if, $x^3=y^3$, this implies that
$$(x-y)(x^2+xy+y^2)=0$$
Now, we 'note' that
$$x^2+xy+y^2=\left(x+\frac{y}{2}\right)^2+\frac{3}{4}y^2$$
Thus if the product above is zero, we must have that either
$$x-y=0\qquad\text{or}\qquad\left(x+\frac{y}{2}\right)^2+\frac{3}{4}y^2=0$$
and this latter can only happen if $x=y=0$. Thus $x=y$ in either case, and $f(x)$ is injective.
I wondered if this approach could tackle $g(x)=x^5$. And, indeed, $x^5=y^5$ implies
$$(x-y)\left(\frac{3}{4}\left[x^2-\frac{1}{2}y^2\right]^2+\frac{1}{4}\left[x+y\right]^4+\frac{2}{3}y^4\right)=0$$
from which we see $g(x)$ is injective. Clearly, this method is going to become extremely cumbersome as the power increases, and we'd like to have elementary calculus to show that $h(x)=x^{2n+1}$ is injective on all of $\mathbb{R}$. But, just for entertainment, I'd like to see
$$x^6+x^5y+x^4y^2+x^3y^3+x^2y^4+xy^5+y^6$$
written as a sum of even-degree terms (if it can) to give an elementary proof - if retractable proof - that $k(x)=x^7$ is indeed injective. How one approached the problem would also be insightful. My attempts haven't been successful.
Any help is appreciated.
 A: There is a simple recursive reduction, at least for $7$.
Write:
$$\begin{align}
x^6+x^5+x^4+x^3+x^2+x+1 &= \left(x^3+\frac{1}{2}x^2\right)^2 + \frac{3}{4}x^4+x^3+x^2+x+1\\
&=\left(x^3+\frac{1}{2}x^2\right)^2 + \frac{3}{4}\left(x^2+\frac{2}{3}x\right)^2 + \frac{2}{3}x^2+x+1\\
&=\left(x^3+\frac{1}{2}x^2\right)^2 + \frac{3}{4}\left(x^2+\frac{2}{3}x\right)^2+\frac{2}{3}\left(x+\frac{3}{4}\right)^2 + \frac{5}{8}
\end{align}
$$
Basically, the most obvious induction works here. If $p_n(x)=\frac{x^{2n+1}-1}{x-1}$ then 

Claim: $p_n(x)$ can be written as a sum of squares with one constant term $\alpha_n$ in the range $(1/2,1]$. 

Then prove inductively that since:
$$p_{n+1}(x) = x^2p_n(x) + x + 1$$
And we can just write:
$$\alpha_n x^2 + x+1 = \alpha_n\left(x+\frac{1}{2\alpha_n}\right)^2 + \left(1-\frac{1}{4\alpha_n}\right)$$
Since $\alpha_n\in(1/2,1]$ implies $1-\frac{1}{4\alpha_n}\in (1/2,1]$, we are done. (Well, technically, we need to show true for $n=1$.)
There is a closed form for $\alpha_n$:
$$\alpha_n = \frac{n+2}{2(n+1)} = \frac{1}{2} + \frac{1}{2(n+1)}$$
Since $\alpha_n$ is also used to determine the terms, that is $$\frac{1}{2\alpha_n} = \frac{n+1}{n+2}$, we can get a closed form:

$$1+x+\dots x^{2n} = \frac{1}{2}+\frac{1}{2(n+1)} + \sum_{k=0}^{n-1} \frac{k+2}{2(k+1)}\left(x^{n-k}+ \frac{k+1}{k+2}x^{n-k-1}\right)^2$$

If you must get the result back for two variables, we write $y^{2n}p_n(x/y)$, you get:
$$\frac{x^{2n+1}-y^{2n+1}}{x-y} =\left( \frac{1}{2}+\frac{1}{2(n+1)}\right)y^{2n} + \sum_{k=0}^{n-1} \frac{k+2}{2(k+1)}\left(x^{n-k}y^k+ \frac{k+1}{k+2}x^{n-k-1}y^{k+1}\right)^2$$

An alternative question is whether you can always write it as as a symmetric sum in $x,y$. For example, solving: $y^2+xy+x^2=(ax+by)^2+(bx+ay)^2$ yields $a,b=\frac{\sqrt{2\pm\sqrt{3}}}2$.
We can fudge with $2n$ terms by taking $\frac{f(x,y)+f(y,x)}{2}$ using our original formula, but is that the best we can do, or can we find $n+1$ terms that give us this result so that the terms are "obviously" symmetric, as we see with $n=1$ even, we get ugly results constants.
