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When is (m+1)/(m-1) an integer?

How do you prove it? m is an integer >1.

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Is the term which should be proved a quotient? If yes the variable m may not be 1, otherwise m-1 in denominator is 0.
(edited 9 years ago)
Reply 2
Original post by Kallisto
Is the term which should be proved a quotient? If yes the variable m may not be 1, otherwise m-1 in denomator is 0.
What do you mean? I stated m is an integer greater than 1.
Reply 3
Original post by NoNewFriends
How do you prove it? m is an integer >1.

m+1m1=m1+2m1=m1m1+2m1=...\displaystyle \frac{m+1}{m-1} = \frac{m-1+2}{m-1} = \frac{m-1}{m-1} + \frac{2}{m-1} = ...

Does that help?
Original post by NoNewFriends
What do you mean? I stated m is an integer greater than 1.


Never mind! I have already seen that its a quotient. Well, well. I have found by trying out that the term is integer for m = -1, 0, 2, 3.
Reply 5
Original post by notnek
m+1m1=m1+2m1=m1m1+2m1=...\displaystyle \frac{m+1}{m-1} = \frac{m-1+2}{m-1} = \frac{m-1}{m-1} + \frac{2}{m-1} = ...

Does that help?
I wrote it like that but no luck. Is 2/(m-1) <1 for m>3? So the expression is 1+r where r is not an integer, so can't be integer.
Reply 6
Original post by NoNewFriends
I wrote it like that but no luck. Is 2/(m-1) <1 for m>3? So the expression is 1+r where r is not an integer, so can't be integer.

Why no luck?

What you've written looks fine to me.

Edit: You should probably say 0<2/(m-1)<1 instead of just 2/(m-1)<1.
(edited 9 years ago)
Reply 7
Original post by notnek
Why no luck?

What you've written looks fine to me.

Edit: You should probably say 0<2/(m-1)<1 instead of just 2/(m-1)<1.
Thank you very much.
Original post by notnek
m+1m1=m1+2m1=m1m1+2m1=...\displaystyle \frac{m+1}{m-1} = \frac{m-1+2}{m-1} = \frac{m-1}{m-1} + \frac{2}{m-1} = ...

Does that help?


I see. That is a nice change. The term m1m1\displaystyle \frac{m-1}{m-1} is always the same for every value m (except m = 1), as numerator and denominator are the same. So to get an integer number, the term 2m1\displaystyle \frac{2}{m-1} must be an integer number itself. Very clever! :smile: that is to say whenever 2m1\displaystyle \frac{2}{m-1} is an integer number, the whole sum is an integer number too, isn't it?
(edited 9 years ago)
Reply 9
Original post by Kallisto
that is to say whenever 2m1\displaystyle \frac{2}{m-1} is an integer number, the whole sum is an integer number too, isn't it?

That's right. And the fraction can only be an integer when m12m-1\leq 2 i.e. m3m\leq 3
Original post by notnek
That's right. And the fraction can only be an integer when m12m-1\leq 2 i.e. m3m\leq 3


Right. My calculation confirmed that, as I have not found another values out which fits to the term. So from -1 to 3 all values (except 1, because its not definable) come into question.
How do you find all positive integers n such that n+1|n-1?
(edited 9 years ago)
Reply 12
Original post by NoNewFriends
How do you find all positive integers n such that n+1|n-1?

There aren't any. Are you sure you've written the question correctly?
Original post by Kallisto
The term m1m1\displaystyle \frac{m-1}{m-1} is always the same for every value m (except m = 1), as numerator and denominator are the same.


You can just write this as 1, therefore it shouldn't be involved in determining whether the result is an integer. (Also, the question states m>1)
Original post by notnek
There aren't any. Are you sure you've written the question correctly?
It is fine that there aren't any. How do you prove it though?
Original post by NoNewFriends
It is fine that there aren't any. How do you prove it though?


Is the change in comment #4, the values in #5, the explanations in #9 and the conclusion in #10 not enough to state a proof?
Reply 16
Original post by Kallisto
Is the change in comment #4, the values in #5, the explanations in #9 and the conclusion in #10 not enough to state a proof?

That would be enough to prove values of n for n-1|n+1 but not the other way around.
Reply 17
Original post by NoNewFriends
It is fine that there aren't any. How do you prove it though?

Well why aren't there any integers? Convince yourself and then you can prove it.

I say "prove" but you don't really need to write much here to explain that there are no such integers.
Original post by notnek
Well why aren't there any integers? Convince yourself and then you can prove it.

I say "prove" but you don't really need to write much here to explain that there are no such integers.
I can convince myself and can prove it using what you helped me prove earlier (with sign adjustment). I thought there would be a simpler way.
Is the explanation n+1|n-1 implies n+1 <= n-1 which doesn't make sense as n is positive integer?

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