Home » Ask & Discuss » Mathematics. » Algebra « Back to Discussion

Algebra

Forum Expert

Joined:	28 Feb 2007
Post:	2173

19 Mar 2008 18:33:33 IST

0 People liked this

645

Number Theory

None

I expect the seniors to be busy mugging for their exam tomorrow. So continue doing that.

Meanwhile those who are free pls do this

Prove that if m is a natural number m(m+1) can never be a perfect power i.e. not a square or a cube etc. of a natural number.

Share this article on:

Comments (13)

S Chandravadan

Hot goIITian

Joined: 25 Jan 2007

Posts: 159

19 Mar 2008 18:35:59 IST

Like 0 people liked this

Well, is it because no natural number can divide two consecutive numbers exactly? (1 an exception, of course)

Hari Shankar

Forum Expert

Joined: 28 Feb 2007

Posts: 2173

19 Mar 2008 18:37:34 IST

Like 0 people liked this

I cant answer that as I would be giving away the answer. Just give me a rigorous proof.

sandeep ramesh

Blazing goIITian

Joined: 13 Mar 2008

Posts: 1183

19 Mar 2008 18:47:25 IST

Like 0 people liked this

sir im not mugging :D :D

sandeep ramesh

Blazing goIITian

Joined: 13 Mar 2008

Posts: 1183

19 Mar 2008 18:48:50 IST

Like 1 people liked this

proof for perfect square is simple

consider m^2 and (m+1)^2, the given term is between that :D

Abhijith

Blazing goIITian

Joined: 23 Jan 2007

Posts: 678

19 Mar 2008 18:51:37 IST

Like 1 people liked this

proof: GCF of 'm' and 'm+1' is 1!

sir did u mean "the product of 'n' consecutive numbers can never be a number to the power of another number"? that would be a really nice problem. someone try doing it when they get time.

btw, i am not mugging either!

sandeep ramesh

Blazing goIITian

Joined: 13 Mar 2008

Posts: 1183

19 Mar 2008 18:53:15 IST

Like 0 people liked this

yes nice, thus meaning that both of them have to be perfect nth powers :)

sandeep ramesh

Blazing goIITian

Joined: 13 Mar 2008

Posts: 1183

19 Mar 2008 18:53:39 IST

Like 0 people liked this

which is clearly not possible :)

Abhijith

Blazing goIITian

Joined: 23 Jan 2007

Posts: 678

19 Mar 2008 18:54:04 IST

Like 1 people liked this

exactly!

sandeep ramesh

Blazing goIITian

Joined: 13 Mar 2008

Posts: 1183

19 Mar 2008 18:58:42 IST

Like 0 people liked this

brilliant man :D Hats off :)

i was thinking another way like bcos one of them is even, it shd be in the form of 2^n * k for it to be a nth power :) implyinh that either its successor shd be a nth power of 3 or sth :) in one of the cases

Hari Shankar

Forum Expert

Joined: 28 Feb 2007

Posts: 2173

19 Mar 2008 19:03:41 IST

Like 0 people liked this

i dont like this at all. ur parents r gonna be mad.

but pls post a full and complete solution.

sandeep ramesh

Blazing goIITian

Joined: 13 Mar 2008

Posts: 1183

19 Mar 2008 19:07:25 IST

Like 0 people liked this

my parents arent here :D

Well consider what koni said, the 2 nos have gcd of 1, meaning that both should be perfect nth powers but clearly 2 adjacent numbers cant be nth powers as the smallest diff between 2 squares is 3 for n belongs to N and x^n is more increasing fn than x^2 so QED :)

Abhijith

Blazing goIITian

Joined: 23 Jan 2007

Posts: 678

19 Mar 2008 19:11:40 IST

Like 1 people liked this

ok here is more explanation.

Let us assume for two consecutive integers the product is a power of another number 'k'.

Hence,

For this to be true the numbers should be of this form:

such that

But that would imply that they have a common factor of 'k'.

But this is not possible since the greatest common factor of any 2 consecutive integers is 1.

Hence this is not possible.

and my parents are sleeping!

vineetnegi

Hot goIITian

Joined: 26 Feb 2008

Posts: 117

19 Mar 2008 20:49:24 IST

Like 0 people liked this

m(m+1) will always be even
odd numbers powers are odd (just a thought)

Quick Reply

Reply

	Some HTML allowed.
	Keep your comments above the belt or risk having them deleted.
	Signup for a avatar to have your pictures show up by your comment
	If Members see a thread that violates the Posting Rules, bring it to the attention of the Moderator Team