Algebra

hemang's Avatar
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Joined: 27 Dec 2010
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29 Feb 2012 15:27:06 IST
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simple question.
Engineering Entrance , JEE Main , JEE Main & Advanced , Mathematics , Algebra

prove that for any positive ineteger n,

n(n+1)(n+2)(n+3)(n+4)(n+5) can never be a perfect square. this is an IMO question! well indirectly. i have changed it a bit. try it.

 



Comments (6)

prahlad kumar sharma's Avatar

Cool goIITian

Joined: 21 Jul 2010
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1 Mar 2012 10:20:34 IST
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putting n = 1,2,3,4 and 5 we get that the resulting number is nt perfect square

so here's the proof for n greater than 5

1st observe that for any n greater than 5, atleast one prime no.(greater than 5) exists betwwen any 6 consecutive intergers

so atleast one of n,(n+1),(n+2),(n+3),(n+4),(n+5) is prime

1st case:(when n is prime)

n is prime so (n+1)(n+2)(n+3)(n+4)(n+5) must have a common factor of n so that n(n+1)(n+2)(n+3)(n+4)(n+5) can be a perfect square

now in expansion of (n+1)(n+2)(n+3)(n+4)(n+5) every term will contain n except 5*4*3*2*1

and 5*4*3*2*1 = 235 but the prime  n is greater than 5 so we cant take common from 235

2nd case: (when n+1 is prime)

let n+1=x

=> n = x-1  so (n+1)(n+2)(n+3)(n+4)(n+5) becomes (x-1)(x)(x+1)(x+2)(x+3)(x+4)

since x is prime (x-1)(x+1)(x+2)(x+3)(x+4) must contain a common factor x so that (x-1)(x)(x+1)(x+2)(x+3)(x+4) can be a perfect square

in exp. of (x-1)(x+1)(x+2)(x+3)(x+4) every term contains x except 4*3*2*1 and 4*3*2*1 = 23*3 but the prime x is greater than 5 so here also we cant take common

similarly going for the rest case we can prove that n(n+1)(n+2)(n+3)(n+4)(n+5) can never be a perfect square.

 

 

 

 

hemang's Avatar

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Joined: 27 Dec 2010
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1 Mar 2012 15:12:31 IST
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you said that at least one prime exists between any 6 consecutive integers greater than 5.

wrong..

primes are not distributed like that.

for example i can name many consecutive numbers all of which are composite.

like 1001! + 2, 1001! + 3, , .... , 1001! + 1001.

each of these is composite ( can you see why???)

so clearly your observation doesn't have enough support.

keep thinking!!! 

prahlad kumar sharma's Avatar

Cool goIITian

Joined: 21 Jul 2010
Posts: 88
1 Mar 2012 20:19:50 IST
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oh sorry sorry

i dont have enogh time to think now , exams gonna start 2morrow but a hint would be this

for two consecutive intergers n,n+1

n2<n(n+1)<(n+1)    so product of 2 consecutive intergers cant be a perfect square

also, n(n+1)(n+2)(n+3) = (k2+3k+1)-1  so product of 4 consecutive intergers cant be a perfect sqaure.

rest i will do 2morrow

prahlad kumar sharma's Avatar

Cool goIITian

Joined: 21 Jul 2010
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1 Mar 2012 20:23:15 IST
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*typing mistake

n2<n(n+1)<(n+1)2

n(n+1)(n+2)(n+3) = (k2+3k+1)2-1

 

prahlad kumar sharma's Avatar

Cool goIITian

Joined: 21 Jul 2010
Posts: 88
1 Mar 2012 20:24:52 IST
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*typing mistake

n2<n(n+1)<(n+1)2

n(n+1)(n+2)(n+3) = (k2+3k+1)2-1

 

hemang's Avatar

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Joined: 27 Dec 2010
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2 Mar 2012 11:27:28 IST
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hmmm.

i actually didn't get this part -

n(n+1)(n+2)(n+3)  = (k^2 + 3k + 1)^2 - 1.

shouldn't it have been n in place of k in the rhs?

its a fundamental result.

n(n+1)(n+2)(n+3) = (n^2 + 3n + 1)^2 - 1.

nice thought process. let us see where it can get us.

 




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