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    (Original post by jack.hadamard)
    Cassini's identity is regularly used in STEP. You are meant to recognise that and prove no other solutions.
    Oh that thing! I've never been good at remembering names
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    (Original post by Jkn)
    x
    Notational question-what does it mean to take the product from 1 to 0? i.e. if k=0 what does it mean?
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    (Original post by Jkn)
    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 )
    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.

    (Original post by Jkn)
    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.
    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?


    (Original post by jack.hadamard)
    You are on the team this year, then? It's approaching.
    Nope, I was 2 points below the sixth.
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    Solution 234*

    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.
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    (Original post by bananarama2)
    Congrats btw
    I'm now constantly worrying that it wasn't good enough to pass :banghead:
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    (Original post by james22)
    Notational question-what does it mean to take the product from 1 to 0? i.e. if k=0 what does it mean?
    Appologies! I was trying to be careful is my definition but must've made a typo. K ranges from 1 to n-1
    (Original post by Mladenov)
    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.
    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?
    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?
    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?
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    (Original post by ukdragon37)
    I'm now constantly worrying that it wasn't good enough to pass :banghead:
    Well you can join the rest of us in that now
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    (Original post by james22)
    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
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    (Original post by Jkn)
    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
    Not sure what the error was, how the question looks now is how I interpreted it.
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    (Original post by Jkn)
    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?
    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}

    (Original post by Jkn)
    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?
    We have also the relation (m,n) \mapsto (n-m,m). I can't get your point.
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    (Original post by Mladenov)
    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.
    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?!
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    (Original post by Mladenov)
    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}
    But there's no "i" in either of up the upper limits of you sigma expression... ?
    We have also the relation (m,n) \mapsto (n-m,m). I can't get your point.
    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...
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    Guys, may i just say, that it's really beautiful to witness so many young minds inventing precocious proofs. Although I'm not too deeply submerged into the realm of extra curricular problem solving, I'm able to assimilate the degree of talent some of you guys are exhibiting. It's scintillating. Just scintillating.
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    (Original post by james22)
    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?
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    (Original post by Jkn)
    Hmm I'm confused then :/

    Do you get x_1-\frac{1}{x_1} =1 for the first part?
    Just realised I messed up the algebra, the equation actually does have real roots.
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    (Original post by Jkn)
    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?!
    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?!

    (Original post by Jkn)
    But there's no "i" in either of up the upper limits of you sigma expression... ?
    I can't. Look again at the upper limits of each sum, and check my solution once again.

    (Original post by Jkn)
    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...
    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.
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    (Original post by Jkn)
    Hmm I'm confused then :/
    That's my permanent state of mind at the minute.
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    After the exams are over, If I decide not to work over the summer I'll be enjoying spending time on this thread.
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    (Original post by Mladenov)
    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?!
    Yes but if its an Olympiad inequalities question then the idea is that you show the use of olympiad inequalities rather than the trivial algebraic rearrangement at the end.

    Did everyone thats reading this instantly 'assume' the following steps in their heads whilst reading the proof?

    By \frac{1}{GM} \ge \frac{1}{AM}, \dfrac{1}{(n-1) \sqrt[n-1]{(n-1)!} \sqrt[n]{n!}} \ge \dfrac{1}{(n-1)\left(\frac{1}{n-1}(1+2+...+n-1) \right) \left(\frac{1}{n}(1+2+...+n) \right)} 

= \dfrac{2}{n(n+1)(n+2)}
    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.
    Well if your contradiction meant that the solution was non-trivial, why didn't you... find the solution?
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    Guys, let's leave the acerbity to PMs. Peace.
 
 
 
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