
#1
Dec2412, 07:50 AM

P: 135

Question is :
If m,k,n are natural numbers and n>1, prove that we cannot have m(m+1)=k^{n}. My attempt : Using induction: If m follows this rule... m+1 must follow it .. so (m+1)(m+2) = k^{n} Since every natural number can be expressed as the product of primes, it follows that (m+1) and (m+2) are primes. Now, my question is... are there any two consecutive primes which can be expressed in the form of k^{n}. Thanks for any help. 



#2
Dec2412, 08:31 AM

Sci Advisor
P: 778

Also where have you used the induction step? 



#3
Dec2412, 12:44 PM

P: 891





#4
Dec2412, 09:56 PM

P: 1,351

a mathematics olympiad problem
hint: m and (m+1) are relatively prime, so any prime factor of k, divides either m or m+1




#5
Dec2412, 11:32 PM

P: 135

Is that enough or needs a better proof ? Thanks for the hint .... 



#6
Dec2512, 03:41 AM

P: 1,351

Spoiler
This means that if a prime factor p of k divides m, then p^n has to divide m, so m has to be an n'th power, and so has m+1, but different n'th powers differ by more than one if m>0, so m and m+1 can't be both n'th powers, but we just proved they have to be, producing a contradiction




#7
Dec2512, 04:08 PM

P: 25

The thing to notice is that m and m+1 are relatively prime. This means that the only factor of k which can divide m and m+1 simultaneously is 1. Hence both m and m+1 must be nth powers. This is impossible for n>1.




#8
Dec2612, 03:03 AM

P: 135

I have found a solution to the problem. It was a bit difficult to type, so I wrote it and attached two .jpg files. Let me know if it's correct ...




#9
Dec2612, 05:08 AM

P: 1,351





#10
Dec2612, 05:50 AM

P: 135

I request you to consider it once again. Let me be clear with it ...
If m(m+1) = k^n, then m(m+1)/k*k ... n = 1 alright upto here ? Then it's clear that k must divide either m or m+1 but it can't divide both. I don't believe there's anything wrong with it. I would like to seek opinion of others (mathsman1963,ramsey) also. If they say it's not upto mark, then I would pick some other step. Thanks. 



#11
Dec2612, 08:19 AM

Sci Advisor
P: 778





#12
Dec2612, 08:30 AM

P: 891

P.S. problems in number theory are interesting in that the details can be put into specific criteria and that there are facts that if properly associated with specific criteria will allow for solutions to more complicated problems. Some problems have proven to be seemingly imposible to solve, but that only makes it more interesting. You have a interest in math. This is good since the world need good mathematicans. In number theory, products of primes, squares, congruences, and powers of numbers are basic building blocks. Once you understand those, there are a whole world of interesting concepts to explorer. 



#13
Dec2612, 10:11 AM

P: 135

I now understand what you people wanna say .... but I don't find any way to solve the problem that way ...
For example I considered the number 6 = 2 * 3. Simply, I can say that 2(3) when divided by k = 6 is divisible by k as n = 1 here but k = (6)^2 can't divide 2(3). How do I put it in variables. I am new to number theory. Please give a solution this time ... I would certainly try some other on my own some day. 



#14
Dec2612, 12:35 PM

P: 891





#15
Dec2612, 09:36 PM

Sci Advisor
HW Helper
P: 9,422

say n = 2. then m and m+1 are both squares, which is not likely.




#16
Dec2712, 03:17 AM

P: 135

I got it. Thanks a lot for teaching me something new. Such fabulous proofs of number theory make me a fan of it. I wanna master it but don't know where to begin. Please suggest me some books. Thanks again ...




#17
Dec2712, 12:16 PM

Sci Advisor
HW Helper
P: 9,422

there are many good books on number theory at a wide variety of prices. here is a good elementary one cheap.
http://www.amazon.com/ElementaryNum...+number+theory 


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