Two little problems (ii)

Problem 2. Find the solutions of $x^6 + y^6 + z^6 = 6xyz - 3$.

Equalities like this ($xyz$? Sixth powers? I don't know the formula for that!) can look intimidating, but a cool head and some experience will make dealing with such unfamiliar things a bit more familiar.

Tempted to simplify the right hand side, we divide by three:
\[
\frac{x^6 + y^6 + z^6}{3} = 2xyz - 1
\]
The key of the exercise is to notice how the left hand side can be seen as an arithmetic average of 3 values, $x^6$, $y^6$ and $z^6$. Given how either the arithmetic average is always greater than the geometric average, or all the values are equal, we can consider the inequality:
\[
\sqrt[3]{x^6y^6z^6} \leq \frac{x^6 + y^6 + z^6}{3}
\]
And we simplify:
\[
x^2y^2z^2 \leq \frac{x^6 + y^6 + z^6}{3}
\]
We now define $xyz = t$, and proceed to exchange the fraction by the equivalence we first saw:
\[
t^2 \leq 2t - 1
\]
Therefore $(t-1)^2 \leq 0$, which is only possible for $t=1$ (since any other value will result in a positive square).

Now, since the product of the three variables is 1, $\sqrt[3]{x^6y^6z^6} = \sqrt[3]{1^6} = 1$, and going back to the initial equality, $2xyz - 1 = 2 - 1 = 1$, which is equal as well to the arithmetic average, we find that both the $AM$ and the $GM$ are equal, and therefore their elements are equal as well. Then:
\[
x^6 + x^6 + x^6 = 3x^6 = 3
\]
And so $x^6 = 1$. We find six solutions for $x$, all of them being either 1 or -1, but we have to restrict to combinations such that $xyz = 1$, which is left as a petty exercise for the reader.

We did this problem in class, and everyone was quite fascinated with how the problem simply unraveled with the use of such elementary methods. Admittedly, it's a very synthetic problem, but it testes your ability to see that which is not immediately obvious to the eye.

Problem 3. Any power $n^k$ can be expressed as the sum of $n$ consecutive odd integers.

This one comes from proving the $2^k$ case, and then being asked to do the same with $3^k$. It seemed too good to be true, but as it turns out, the proof is quite simple.

We assert the theorem:
\[
n^k = (2q + 1) + (2q + 3) + (2q + 5) + ... + (2q + (2n - 1))
\]
Let's reorder this:
\[
n^k = n2q + (1 + 3 + 5 + 7 + ... (2n - 1))
\]
It is a fact that the sum of the first $r$ odds is equal to $r^2$ (it's a one-line proof), so:
\[
n^k = n2q + n^2
\]
We find $q$:
\[
q = \frac{n^{k-1} - n}{2}
\]
And then verify the identity:
\[
n^k = n2\left(\frac{n^{k-1} - n}{2}\right) + n^2 = n(n^{k-1} - n) = n^n - n^2 + n^2 = n^k
\]
And we're done; the identity derived from the sum of $n$ consecutive odd integers is proved to be equal to $n^k$!