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    (Original post by Zephyr1011)

    Solution 88

    Assume that a_k\geq5. Then a_k-2\geq3, and so a different combination of integers, where a_k is replaced with a_k-2 and 2, will have a greater product as 2(a_k-2)=(a_k-2)+(a_k-2)>(a_k-2)+2=a_k. Therefore a_k<5 \;\forall k
    Nice and rigorous
    If a_k=4, this is equivalent to a combination where a_k is replaced with 2 2s, as they have the same product and sum. So, WLOG, there must be an optimal combination with a_k<4\;\forall k. Clearly if a_k=1, a combination where another number was instead increased by 1 would have a greater product, therefore a_k\neq1. Therefore a_k=2 or 3. If there are 3 or more terms which are 2, then a combination where there are 2 3s, instead of 3 of those 2s, there will be a greater product as 2+2+2=3+3 and 2^3<3^2. Therefore the optimal combination will have only 3s and 2s, and at most 2 2s.

    Therefore, if N=0 \pmod{3}, P(N)=3^{\frac{N}{3}}, if N=1 \pmod{3}, P(N)=4\cdot3^{\frac{N-4}{3}} and if N=2 \pmod{3}, P(N)=2\cdot3^{\frac{N-2}{3}}
    You've used the same method as mladenov, but I do find that the answer, when written in modular arithmetic form, is more aesthetically pleasing.

    Do share some problems!
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    (Original post by Jkn)
    You've used the same method as mladenov, but I do find that the answer, when written in modular arithmetic form, is more aesthetically pleasing.

    Do share some problems!
    Oh, sorry, I hadn't seen their solution.

    I will now see if I can find any interesting problems.

    Edit: Ah, found one. The only problem I managed to get in the BMO2

    Problem 93*

    Prove there are an infinite number of positive integers m and n, such that m|n^2+1 and n|m^2+1
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    Problem 94**

    One really nice functional equation.

    Find all strictly increasing functions f: \mathbb{R} \to \mathbb{R} such that f(x+f(y))=f(x+y)+1, for all real numbers x and y.

    Problem 95** warm-up

    Let a,b,c be positive real numbers. Then, \displaystyle \sum_{cyc} \frac{a^{2}b^{2}(b-c)}{a+b} \ge 0.

    Problem 93 has been already posted.
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    (Original post by Zephyr1011)
    Oh, sorry, I hadn't seen their solution.

    I will now see if I can find any interesting problems.
    What I've been doing in some cases is trying to think of some myself; often generalisations of other problems I have tried/found Such as...

    Problem 96*

    Let x, y and a be positive integers such that x^3+ax^2=y^2. Given that  1 \leq x \leq b where b is a positive integer, find, in terms of a and b, the number of possible pairs (x,y) that satisfy the equation.
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    (Original post by Zephyr1011)
    Prove that \forall n\in\mathbb{N}, there exists n consecutive integers such that none of them are primes or a power of a prime.
    Problem 8 I do believe
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    We have totally messed up the numbers!

    Solution 93

    Let F_{n} be the nth Fibonacci number. From Catalan's identity, we have F_{2n+1}^{2}+1=F_{2n-1}F_{2n+3}, and F_{2n-1}^{2}+1=F_{2n-3}F_{2n+1}. Hence F_{2n+1}^{2}+1 \equiv 0 \pmod {F_{2n-1}} and F_{2n-1}^{2}+1 \equiv 0 \pmod {F_{2n+1}}.
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    (Original post by Mladenov)
    Solution 93

    Let F_{n} be the nth Fibonacci number. From Catalan's identity, we have F_{2n+1}^{2}+1=F_{2n-1}F_{2n+3}, and F_{2n-1}^{2}+1=F_{2n-3}F_{2n+1}. Hence F_{2n+1}^{2}+1 \equiv 0 \pmod {F_{2n-1}} and F_{2n-1}^{2}+1 \equiv 0 \pmod {F_{2n+1}}.
    HAHAHAHAHAHAHAHAHAHA :lol: absolute legend! :lol: That took you less than a minute! :lol:
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    (Original post by Mladenov)
    Problem 94**

    One really nice functional equation.

    Find all strictly increasing functions f: \mathbb{R} \to \mathbb{R} such that f(x+f(y))=f(x+y)+1, for all real numbers x and y.
    If x=0, then f(f(y))=f(y)+1. Therefore, for all values of a in the range of f, f(a)=a+1

    This works as f(x+f(y))=x+y+2=f(x+y)+1 and f is clearly a strictly increasing function

    Does the question state that the range of f is all reals, or a subset of all reals?
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    (Original post by Jkn)
    HAHAHAHAHAHAHAHAHAHA :lol: absolute legend! :lol: That took you less than a minute! :lol:
    Number theory


    (Original post by Zephyr1011)
    If x=0, then f(f(y))=f(y)+1. Therefore, for all values of a in the range of f, f(a)=a+1

    This works as f(x+f(y))=x+y+2=f(x+y)+1 and f is clearly a strictly increasing function

    Does the question state that the range of f is all reals, or a subset of all reals?
    Yup, the range of f is the set of all real numbers.
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    (Original post by Mladenov)
    We have totally messed up the numbers!

    Solution 93

    Let F_{n} be the nth Fibonacci number. From Catalan's identity, we have F_{2n+1}^{2}+1=F_{2n-1}F_{2n+3}, and F_{2n-1}^{2}+1=F_{2n-3}F_{2n+1}. Hence F_{2n+1}^{2}+1 \equiv 0 \pmod {F_{2n-1}} and F_{2n-1}^{2}+1 \equiv 0 \pmod {F_{2n+1}}.
    ...That question took me 2 hours.

    I mean, actually solving the question only took up around a quarter of that time, most of the rest was spent messing up the proof by induction, which I'd used to solve it and I've never heard of Catalan's Identity, but still...

    Oh, and should I put the old Problem 93 back now, since there's a solution to it?
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    (Original post by Zephyr1011)
    If x=0, then f(f(y))=f(y)+1. Therefore, for all values of a in the range of f, f(a)=a+1

    This works as f(x+f(y))=x+y+2=f(x+y)+1 and f is clearly a strictly increasing function

    Does the question state that the range of f is all reals, or a subset of all reals?
    On a side note, are you really 14? :shock:
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    (Original post by Mladenov)
    Yup, the range of f is the set of all real numbers.
    Great, then I think I've proved that \forall a\in\mathbb{R}, f(a)=a+1
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    (Original post by Felix Felicis)
    On a side note, are you really 14? :shock:
    ...Yes
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    (Original post by Zephyr1011)
    Great, then I think I've proved that \forall a\in\mathbb{R}, f(a)=a+1
    You have proved that f(x)=x+1 only on the curve x=0 (not f my bad!), not over the whole complex plane xOy.
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    (Original post by Zephyr1011)
    ...Yes
    Mother of god... :O

    (Original post by Zephyr1011)
    ...That question took me 2 hours.

    I mean, actually solving the question only took up around a quarter of that time, most of the rest was spent messing up the proof by induction, which I'd used to solve it and I've never heard of Catalan's Identity, but still...
    Did you use the exact same method (i.e. fibionacci)? I may try to find an alternative solution (don't give away anything if you did use another method)
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    (Original post by Mladenov)
    You have proved that f(x)=x+1 only on the curve x=0 (not f my bad!), not over the whole complex plane xOy.
    What does xOy mean? And surely since any definition of f must be valid in the case x=0, and my definition is the only one valid in that case, my definition can be the only valid one.
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    (Original post by Jkn)
    Did you use the exact same method (i.e. fibionacci)? I may try to find an alternative solution (don't give away anything if you did use another method)
    Well, I tried some random numbers until I noticed that the only solutions I could find were alternating Fibonnacci numbers, and then proved that as it worked for the first two numbers, it must work for all numbers following it, and that as these terms are always increasing there must be an infinite number of them.
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    (Original post by Zephyr1011)
    What does xOy mean? And surely since any definition of f must be valid in the case x=0, and my definition is the only one valid in that case, my definition can be the only valid one.
    My apologies, I have overlooked your solution. It is correct.

    Here is another one.

    Problem 97**

    Find all continuous functions, defined for every x \in \mathbb{R}, which satisfy f(f(x))=f(x)+x.
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    (Original post by Zephyr1011)
    Well, I tried some random numbers until I noticed that the only solutions I could find were alternating Fibonnacci numbers, and then proved that as it worked for the first two numbers, it must work for all numbers following it, and that as these terms are always increasing there must be an infinite number of them.
    Hmm well in that case I suppose the interesting question here is weather or not there are an infinite number of pairs (m,n) such that m and n are not both Fibonacci numbers?
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    (Original post by Jkn)
    Problem 96*

    Let x, y and a be positive integers such that x^3+ax^2=y^2. Given that  1 \leq x \leq b where b is a positive integer, find, in terms of a and b, the number of possible pairs (x,y) that satisfy the equation.
    Is it not just the number of perfect squares in the interval [a+1,a+b]?
 
 
 
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