Algebra

hemang's Avatar
Blazing goIITian

Joined: 27 Dec 2010
Post: 1508
26 Feb 2012 09:57:21 IST
3 People liked this
4
393 View Post
old is gold.
Engineering Entrance , JEE Main , JEE Main & Advanced , Mathematics , Algebra

this is a famous problem.

prove that the sum of the squares of any 3, 4, 5 or 6 consecutive integers can't be a perfect square.

also give an example of 11 consecutive integers, the sum of whose squares is a perfect square.

 



Comments (4)


New kid on the Block

Joined: 14 Oct 2011
Posts: 4
26 Feb 2012 17:03:36 IST
1 people liked this

 (n-5)2+(n-4)2+(n-3)2+(n-2)2+(n-1)2+(n)2+(n+1)2+(n+2)2+(n+3)2+(n+4)2+(n+5)^2 is a perfect square

=> 11*(n2 + 10) is a perfect square which means n can be 1,-1 

So one possible set of consecutive integers can be -4,-3,-2,-1,0,1,2,3,4,5,6

 

prahlad kumar sharma's Avatar

Cool goIITian

Joined: 21 Jul 2010
Posts: 88
26 Feb 2012 20:39:12 IST
2 people liked this

let me prove for any 3 consecutive integers,rest can de bone in a similar way

 the 3 intergers be (n) , (n+1) , (n+2)

let us 1st assume that sum of the squares these nos. is a perfect square

so,  n2+(n+1)2+(n+2)2=y2

   => 3n2+ 6n+ 5- y2=0

since n is an interger ,the discriminant part of the above quadratic eq. must be a perfect square as to get the value of "n " an interger

so , 36 - 4*3*(5-y2) = 22(y2-6) must be a perfect square but (y2-6) can never be a perfect square

therefore our assumption is wrong.

 

hemang's Avatar

Blazing goIITian

Joined: 27 Dec 2010
Posts: 1508
27 Feb 2012 09:48:57 IST
0 people liked this

hmmm.

11(n^2 + 10) is a perfect square means that 11 divides n^2 + 10.

also, n^2 + 10 is congruent to n^2 - 1 (mod 11).

so, n = 11m + 1 or, 11m - 1.

for some integer m.

now our equation becomes -

11[(11m+or-1)^2 + 10]

or, (11^2)[10m^2 + (m+or-1)^2]

m = 2 works too.

 

hemang's Avatar

Blazing goIITian

Joined: 27 Dec 2010
Posts: 1508
27 Feb 2012 09:57:03 IST
0 people liked this

for prahlad.

to prove that y^2 - 6 can't be a perfect square.

i won't directly use modular algebra. a slightly different path but essentially the same.

alll integral numbers can be wrtten in one of the following ways  -  

4k, 4k+1, 4k+2, 4k+3 .  k is some integer.

so all squares can be written in the following ways -

16k^2 , that is 4m for some integer m.

16k^2 + 8k +1. that is 4p+1 for some integer p.

16k^2 + 4 + 16k. that is 4x for some integer x.

16k^2 + 9 + 24k. that is 4y+1 for some integer y.

so, all perfect squares leave remainders -

1 or 0 on division by 4.

if y^2 leaves the remainder 0, then y^2-6 leaves 2. contradiction.

if y^2 leaves the remainder 1, then y^2 - 6 leaves 3. contradiction.

hence proved.

:)




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
Free Sign Up!
Sponsored Ads

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

Name

Mobile

E-mail

City

Class

Vertical Limit

Top Contributors
All Time This Month Last Week
1. Bipin Dubey
Altitude - 16545 m
Post - 7958
2. Himanshu
Altitude - 10925 m
Post - 3836
3. Hari Shankar
Altitude - 10085 m
Post - 2217
4. edison
Altitude - 10825 m
Post - 7804
5. Sagar Saxena
Altitude - 8635 m
Post - 8064
6. Yagyadutt Mishr..
Altitude - 6330 m
Post - 1958

Find Posts by Topics

Physics

Topics

Mathematics

Chemistry

Biology

Parents Corner

Board

Fun Zone