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IF you would like help with any math problems, feel free to post them

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Original post by Dog4444
I found a good problem, but I can't manage to solve it. It supposed to be pretty easy one:

Prove that there are infinitely many positive integers nn such that n21n^2-1 has a prime divisor greater than 2n+2n 2n+\sqrt{2n}

Help!


I don't think this is true....
Reply 41
Original post by TheMagicMan
I don't think this is true....


Definitely true.
Original post by Dog4444
Definitely true.


n21=(n+1)(n1)n^2-1=(n+1)(n-1) so the greatest prime divisor it can have is n+1n+1. As n+1<2nn+1 < 2n for n>1n >1, there are no numbers for which it is true...
Reply 43
Original post by TheMagicMan
n21=(n+1)(n1)n^2-1=(n+1)(n-1) so the greatest prime divisor it can have is n+1n+1. As n+1<2nn+1 < 2n for n>1n >1, there are no numbers for which it is true...


Holy ****
Typo from me.
It's n2+1n^2+1
Thousand apologises.
Original post by DFranklin
You are reminded to read the Guide to Posting, and in particular what it has to say about posting full solutions.


Ohh shut up man..seriously..this guy has offered his time, skills and experience in helping the students with maths which they are struggling to complete. A step by step guidance which s/he is actually giving, as well as the answer, personally is also helping me. What's the point in posting how to work something out, and not giving the answer for it...it's called knowing if the student will be right or not? Jesus christ...

I bet if you were not a moderator and such ignorant on someones offer i.e. the OP, you would be asking the same kind of questions with the request of the answers.


@Students - Like if you agree to my post.
Reply 45
Original post by FrozenBeak
Ohh shut up man..seriously..this guy has offered his time, skills and experience in helping the students with maths which they are struggling to complete. A step by step guidance which s/he is actually giving, as well as the answer, personally is also helping me. What's the point in posting how to work something out, and not giving the answer for it...it's called knowing if the student will be right or not? Jesus christ...

I bet if you were not a moderator and such ignorant on someones offer i.e. the OP, you would be asking the same kind of questions with the request of the answers.


@Students - Like if you agree to my post.


It would have been better had you conveyed your point by being less rude.

I accept the OP has taken his time to help students. But their are certain flaws of it as well, some students might just give their homework question for the OP to do, and they will then copy without trying to understand it.

DFranklin is a moderator and its his responsibility to point out if there is a rule being broken. He has to follow the forum rules, and ensure nothing against the rules is posted on the forum.

Personally, i really appreciate the effort of the OP.

What's the point in posting how to work something out, and not giving the answer for it


If you are told the method to work something out, then you should use that method to work out the answer, and then tell your answer so that others can check and inform if you are right or wrong.

And IT WILL BE BETTER IF YOU KEEP YOUR OPINIONS TO YOURSELF, SUCH POSTS WON'T HELP ANYONE.
(edited 11 years ago)
Original post by raheem94
It would have been better had you conveyed your point by being less rude.

I accept the OP has taken his time to help students. But their are certain flaws of it as well, some students might just give their homework question for the OP to do, and they will then copy without trying to understand it.

.


Top notch at maths, but terrible English. :tongue: :biggrin:
Reply 47
Original post by Brit_Miller
Top notch at maths, but terrible English. :tongue: :biggrin:


Everybody can make mistakes, i didn't read it well, to eliminate mistakes.
Original post by raheem94
Everybody can make mistakes, i didn't read it well, to eliminate mistakes.


:colone:

Just ribbing, fella.
I was wondering if someone could please help me with part B of following C2 log question attached as I'm not sure how to go about solving it.

Thanks :smile:
Use (a) to substitute the LHS

remove the logs

you have simultaneous equations
Reply 51
How can I prove that this has no solutions for n=3,4,5... where x,y and z are integers?

xn+yn=zn\displaystyle x^n+y^n=z^n
Reply 52
Original post by oh_1993
How can I prove that this has no solutions for n=3,4,5... where x,y and z are integers?

xn+yn=zn\displaystyle x^n+y^n=z^n


Ok, I got it. :colone:
(edited 11 years ago)
Original post by sulexk
Hi,

That is to do with functions of complex variables. I will try and work on it, although have not yet covered Functions of Complex variables. Will try soon though! :smile:
I would love to work on these in the summer, once all exams are done.


Just want to make sure that you know; that is the Riemann Hypothesis and hasn't be solved for over 150 years and counting.

But you could try :wink:
Reply 54
When Sulexk manages to solve all the problems in this thread and becomes famous around the world, we can all look back at this thread and remember that "this was where it all started".
Reply 55
Original post by Dog4444
Ok, I got it. :colone:


:qed: ?
http://www.thestudentroom.co.uk/showthread.php?t=1969325
My q is up here ... please can u help me ??
Hi, working on this q for a while... needhelp please...

A heaving ring of mass 5 kg is threaded on a fixed rough rod. Coefficient of friction between rod and ring is 1/2... A light string is attatched to the string and pulled down at 30 degrees to the horizontal. The magnitude of force T from the light rope is increases from 0. Find the value of T that is just suffiecient to make the equilibrium limiting..

I worked out friction was 24.5 (5x9.8x0.5)
Then I worked out the the horizontal force of the string would be Tcos(30)
So i did
Tcos(30)- 24.5 = 0
Thus, T =28.3 N
The answer is actually 39.8... Where am i going wrong please??
M1 question.jpg
Been stuck on this one all day:

Is it true that every problem whose solution can be quickly verified by a computer can also be quickly solved by a computer?


Spoiler

Reply 59
Original post by oh_1993
How can I prove that this has no solutions for n=3,4,5... where x,y and z are integers?

xn+yn=zn\displaystyle x^n+y^n=z^n


I have an elegant proof but this post is not large enough to contain it.

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