The Proof is Trivial!

Cassini's identity is regularly used in STEP. You are meant to recognise that and prove no other solutions.

x

Seen as it is a problem that gives you a result to prove, you should probably elaborate on your last step (as a rule of thumb, if an IMO examiner would not award you the marks, you can probably assume that we will have trouble following )

This is incorrect. You have assumed a statement without justification and then discarded the 'counterexample' in favour of the conjecture. You have also not fully justified how it is that you have a contradiction and you have also applied an algebraic manipulation that contradicts the generality of a result you later used.

You are on the team this year, then? It's approaching.

Congrats btw

Notational question-what does it mean to take the product from 1 to 0? i.e. if k=0 what does it mean?

I do not know what you mean by "your last step"; all that I use is one combinatorial identity, which can be proved in two lines, and I can do it always.
$\displaystyle \sum_{i=r}^{n} \dbinom{i}{r} = \dbinom{n+1}{r+1}$. There is purely combinatorial proof. Consider the set $\{a_{1}, \cdots, a_{n+1} \}$ of $n+1$ elements. We count the number of $r+1$ element subsets A of this set. Consider the $n-r+1$ cases:
$a_{1} \in A$
$a_{2} \in A$, $a_{1} \not\in A$
$......$
$a_{1}, \cdots a_{n-r-1} \not\in A$, $a_{n-r} \in A$
$a_{1},\cdots,a_{n-r} \not\in A$.
These cases are pairwise disjoint and hence by addition principle the identity follows.

How? It is easy to prove that all the Fibonacci numbers satisfy the required equality, since we have the map $(m,n) \mapsto (n,m+n)$. Amm, $n<m+n$? Which "algebraic manipulation" contradicts to my result?

I'm now constantly worrying that it wasn't good enough to pass

Assume there is a real solution and let k=1, let the product from 2 to n be x so we get 2 simultanious equations in x and x1 which have no real solutions. This is a contradiction.

I wonder if there are any complex solutions? I doubt it but may be harder to prove.

Oh btw, when I changed the k thing there was a rogue n in the wrong place (think this is because I checked through but when my browser crashed, it deleted the corrections along with the other problems) Not sure if you assumed it was an error anyway... but go and have a look and see if it changed anything

When I know how to prove that but you have two sigmas? I just just noticed thoug, one of them is dependant on an i value. Where has it come from?

Well I am confused as to how you made your contradiction then. That step could be applied to any Fibonacci relation, showing that it is clearly not true.

Perhaps observing the existence of trivial integer solutions will convince you that there is no contradiction?

Solution 235

I denote the left handed side by $A_{n}$, for I am too lazy to type this sum. For $n=2$ the inequality holds true. Suppose that it is true for all $i \in \{1,2, \cdots,n\}$.
Then, $\begin{aligned} \displaystyle A_{n+1}> A_{n}+ \frac{2}{n(n+1)(n+2)} = \frac{2n^{3}+6n^{2}+2n-2}{n(n+1)(n+2)} > \frac{2((n+1)^{2}+(n+1)-1)}{(n+1)(n+2)} \end{aligned}$.
It is quite weak.

It is just the previous sum. $\displaystyle \sum_{j=r-1}^{n-1} \dbinom{j}{r-1} +\cdots+ \sum_{j=r-1}^{r-1} \dbinom{j}{r-1}$

We have also the relation $(m,n) \mapsto (n-m,m)$. I can't get your point.

Not sure what the error was, how the question looks now is how I interpreted it.

Hmm I'm confused then :/

Do you get $x_1-\frac{1}{x_1} =1$ for the first part?

Why have you missed all the main steps out in this proof as well? I'm confused.. did you do the hard part using application of AM-GM followed the the sum of squares formula?!

But there's no "i" in either of up the upper limits of you sigma expression... ?

Well I don't understand your point either.

You have said that there is a contradiction and that no values satisfy the equation, right? And yet there are values that satisfy the equation and trivial examples can be found by looking at the Fibonacci sequence.. ? Are you claiming that these values do not exist?! Try (2,1) for example...

Hmm I'm confused then :/

I thought that people here can follow basic steps. It is clear what I have done, as I am using induction on the rest of the terms?!

You are not getting the idea. We suppose that there exists solution which does not belong to the set $T$, and, then, we obtain that it belongs.