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Algebra

Blazing goIITian

 Joined: 27 Dec 2010 Post: 1508
26 Feb 2012 09:57:21 IST
3 People liked this
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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.

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

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.

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.

Blazing goIITian

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

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.

:)

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