Vriti Edustore
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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)
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.
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.
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.
:)
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(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