Home »
Ask & Discuss
»
Mathematics
« Back to Discussion
Algebra
Comments (6)
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!!!
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
n^{2}<n(n+1)<(n+1) so product of 2 consecutive intergers cant be a perfect square
also, n(n+1)(n+2)(n+3) = (k^{2}+3k+1)1 so product of 4 consecutive intergers cant be a perfect sqaure.
rest i will do 2morrow
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.
Take a Free Test  
Scholarship offer: Score above 85% in any of the Free test and get 10% discount on 1 year Test Series  


Preparing for JEE?
Kickstart your preparation with new improved study material  Books & Online Test Series for JEE 2014/ 2015
@ INR 5,443/
For Quick Info
1.  Bipin Dubey

2.  Himanshu

3.  Hari Shankar

4.  edison

5.  Sagar Saxena

6.  Yagyadutt Mishr..

Find Posts by Topics
Physics
TopicsMathematics
Chemistry
Biology
Institutes
Parents Corner
Board
Fun Zone
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 = 2^{3}5 but the prime n is greater than 5 so we cant take common from 2^{3}5
2nd case: (when n+1 is prime)
let n+1=x
=> n = x1 so (n+1)(n+2)(n+3)(n+4)(n+5) becomes (x1)(x)(x+1)(x+2)(x+3)(x+4)
since x is prime (x1)(x+1)(x+2)(x+3)(x+4) must contain a common factor x so that (x1)(x)(x+1)(x+2)(x+3)(x+4) can be a perfect square
in exp. of (x1)(x+1)(x+2)(x+3)(x+4) every term contains x except 4*3*2*1 and 4*3*2*1 = 2^{3}*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.