The Proof is Trivial!

Obviously not, in fact there is no single variable polynomial with that property.

Does this require anything too advanced (like complex analysis)? I'll give it a go but don't want to waste too long on something I will never solve.

Problem 234*

Find all possible n-tuples of reals $x_1,x_2, \cdots ,x_n$ such that $\prod_{i=1}^{n} x_i = 1$ and $\prod_{i=1}^{k} x_i - \prod_{i=k+1}^{n} x_i = 1$ for all $1 \le k \le n-1$

For $1 \leq k \leq n-1$ let $u_k=\prod_{i=1}^{k} x_i$

$1=\prod_{i=1}^{k} x_i - \prod_{i=k+1}^{n} x_i=u_k-\frac{1}{u_k}[br][br]\Rightarrow[br][br]u_{k}^2-u_k-1=0[br][br]\Rightarrow[br][br]u_k=\frac{1 \pm \sqrt{5}}{2}[br][br]\Rightarrow[br][br]x_1=\frac{1 \pm \sqrt{5}}{2}$

We have 2 options for $x_1$, now assume we have found $x_1,x_2,...,x_{m-1}$

Using the 2 possible values of $u_k$ we get 3 options for $x_m$

$x_m=1, \frac{-3+\sqrt 5}{2}, \frac{1+\sqrt 5}{2}$

Where the last 2 are possible depending on the value of $u_{m-1}$

Repeating this process we can get values for $x_1, x_2, ..., x_{n-1}$

This is enough to determine $x_n$ using the first equality so we are done.

By making different choices at each stage in the process used to generate the numbers, we get every possible n-tuple.

Looks good to me bro! (do write solution at the top!)

I assumed this would never have a solution after "Solution 234 Really?!" was posted by a certain someone

It's actually quite ironic given that I got this from the 4th round of the Bulgarian Mathematical Olympiad (which is like our BMO1/BMO2 - perhaps a little easier in places). It's actually a really good source of fun questions that don't follow the typical UK-comp or US-comp format - though it does, in some ways, have a format of it's own! I might try/post a few more actually... (I typed up like 10 back when I set that but the computer froze and deleted the typing that took me ages to do!)

I found it an enjoyable question to think about, although that comment made put me off a bit as I thought there was some really obvious solution I was missing.

To be honest I'm suprised that there were any solutions at all.

Just noticed that this is another place where the golden ratio appears.

It does not require complex analysis. It is on the difficult side, though.

Hmm, odd one - I've made what might be progress, but not very much:
. I'll have another think.

Solution 12

First we choose the six colours, so there are n choose 6 possibilities $=\frac{n!}{6!(n-6)!}$. Without considering identical cases we have $6!$ possibilities for colouring the cube, but we divide by $4*6=24$ where 6 is the number of possible anchor faces and 4 is due to rotation about the anchor face, so we get $\frac{n!}{24(n-6)!}$