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# The Proof is Trivial! Watch

1. (Original post by bananarama2)
Sorry Although I think I may have done you a favor, it took and hour and a half to type that.
It's okay the problem that I have with these questions isn't the theory behind it, it's actually computing it (the amount of mistakes I made first time around was scary )
2. (Original post by cpdavis)
It's okay the problem that I have with these questions isn't the theory behind it, it's actually computing it (the amount of mistakes I made first time around was scary )
Agreed. I missed of loads of terms and was on the verge of giving up, it's distinctly dodgy approximation too.
3. Well done guys!
4. Sorry
5. Solution 53

The proof is trivial for . Now suppose that for some there exists such that .
If . On the other hand, if hence by letting . Noting that it follows inductively that there is always some such that Define . The proposition follows.
6. Problem 60**/***

Evaluate the integral .
7. Solution 48

I have spent about one day on this functional equation, despite the fact that I had done hundreds of functional equations.

We see . Further plugging , we get has three solutions, namely . Hence two of these have to be equal. Suppose, for example, . Then, since , we have . Similarly, by checking the cases and , we conclude that there are no functions such that and .

Now, define , , for all , for , , for , , , , and , for .

More functional equations?

Problem 61**

Find all functions satisfying .
8. (Original post by Star-girl)
Problem 14**

Prove that is divisible by 61.
dont know if anyone has solved this and, ive been unable to find a direct proof - which really annoys me. Heres my best attempt (forgive me please, ive left it here or ill go nuts):

can be factorised using the formula for a^n-1^n.

One of the factors is:

i know! i know! but, i can now unclench my hair and stop grinding my teeth and move on!
9. (Original post by Hasufel)
dont know if anyone has solved this and, ive been unable to find a direct proof - which really annoys me. Heres my best attempt (forgive me please, ive left it here or ill go nuts):

can be factorised using the formula for a^n-1^n.

One of the factors is:

i know! i know! but, i can now unclench my hair and stop grinding my teeth and move on!

There has been a solution that's different to yours but both are equally valid. See the first post in the thread, there is a link to it there (yours will probably be added in soon as well).

The question is a lot easier with a handy little tool called modular arithmetic and even easier of you know a result called Fermat's Little Theorem. You have provided a more accessible solution.
10. (Original post by Star-girl)

There has been a solution that's different to yours but both are equally valid. See the first post in the thread, there is a link to it there (yours will probably be added in soon as well).

The question is a lot easier with a handy little tool called modular arithmetic and even easier of you know a result called Fermat's Little Theorem. You have provided a more accessible solution.
thanks! Didnt see the other solution - will check it out - thank again for the comment!
11. (Original post by Hasufel)
thanks! Didnt see the other solution - will check it out - thank again for the comment!
No problem.
12. (Original post by Star-girl)

You have provided a more accessible solution.
* Better. None of the number theory nonsense
13. (Original post by bananarama2)
* Better. None of the number theory nonsense
This is a thread full of mathematicians... careful what you say.

Also, number theory is pretty awesome.
14. (Original post by Star-girl)
This is a thread full of mathematicians... careful what you say.

Also, number theory is pretty awesome.
I like to live dangerously.

Look at that beauty.

Edit: It won't let me put put the D'alembert symbol.
15. I haven't seen very many inequality questions around, so I'll throw some in.

Problem 62*

Show that if then

Problem 63*

Prove
16. Solution 63

Which is true since a square is equal greater than one.
17. (Original post by bananarama2)
I like to live dangerously.

Look at that beauty.

Edit: It won't let me put put the D'alembert symbol.
Why not use \square
18. (Original post by Dark Lord of Mordor)
Why not use \square
One does not simply use a square.

Spoiler:
Show
That'll do, thanks
19. (Original post by Star-girl)

There has been a solution that's different to yours but both are equally valid. See the first post in the thread, there is a link to it there (yours will probably be added in soon as well).

The question is a lot easier with a handy little tool called modular arithmetic and even easier of you know a result called Fermat's Little Theorem. You have provided a more accessible solution.
I have another solution that I'm quite happy with so I thought I'd post it on here:

The bracket is divisible by 61, as 61 is prime, and divisible by 2 since only 61C1 and 61C60 are odd, due to the otherwise everpresent factor of 60. Hence:

is divisible by 61.
20. (Original post by Star-girl)
I haven't seen very many inequality questions around, so I'll throw some in.

Problem 62*

Show that if then

Problem 63*

Prove
Why does the first one seem so familiar?

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