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# The Proof is Trivial! Watch

1. (Original post by Zephyr1011)

Solution 88

Assume that . Then , and so a different combination of integers, where is replaced with and , will have a greater product as . Therefore
Nice and rigorous
If , this is equivalent to a combination where is replaced with 2 2s, as they have the same product and sum. So, WLOG, there must be an optimal combination with . Clearly if , a combination where another number was instead increased by 1 would have a greater product, therefore . Therefore or . 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 and . Therefore the optimal combination will have only 3s and 2s, and at most 2 2s.

Therefore, if , , if , and if ,
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!
2. (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 and
3. Problem 94**

One really nice functional equation.

Find all strictly increasing functions such that , for all real numbers and .

Problem 95** warm-up

Let be positive real numbers. Then, .

Problem 93 has been already posted.
4. (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 . Given that where b is a positive integer, find, in terms of a and b, the number of possible pairs that satisfy the equation.
5. (Original post by Zephyr1011)
Prove that , there exists n consecutive integers such that none of them are primes or a power of a prime.
Problem 8 I do believe
6. We have totally messed up the numbers!

Solution 93

Let be the th Fibonacci number. From Catalan's identity, we have , and . Hence and .
Solution 93

Let be the th Fibonacci number. From Catalan's identity, we have , and . Hence and .
HAHAHAHAHAHAHAHAHAHA absolute legend! That took you less than a minute!
Problem 94**

One really nice functional equation.

Find all strictly increasing functions such that , for all real numbers and .
If , then . Therefore, for all values of a in the range of f,

This works as 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?
9. (Original post by Jkn)
HAHAHAHAHAHAHAHAHAHA absolute legend! That took you less than a minute!
Number theory

(Original post by Zephyr1011)
If , then . Therefore, for all values of a in the range of f,

This works as 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 is the set of all real numbers.
We have totally messed up the numbers!

Solution 93

Let be the th Fibonacci number. From Catalan's identity, we have , and . Hence and .
...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?
11. (Original post by Zephyr1011)
If , then . Therefore, for all values of a in the range of f,

This works as 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?
Yup, the range of is the set of all real numbers.
Great, then I think I've proved that ,
13. (Original post by Felix Felicis)
On a side note, are you really 14?
...Yes
14. (Original post by Zephyr1011)
Great, then I think I've proved that ,
You have proved that only on the curve (not my bad!), not over the whole complex plane .
15. (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)
You have proved that only on the curve (not my bad!), not over the whole complex plane .
What does mean? And surely since any definition of f must be valid in the case , and my definition is the only one valid in that case, my definition can be the only valid one.
17. (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.
18. (Original post by Zephyr1011)
What does mean? And surely since any definition of f must be valid in the case , 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 , which satisfy .
19. (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 such that m and n are not both Fibonacci numbers?
20. (Original post by Jkn)
Problem 96*

Let x, y and a be positive integers such that . Given that where b is a positive integer, find, in terms of a and b, the number of possible pairs that satisfy the equation.
Is it not just the number of perfect squares in the interval ?

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