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# Prove that the sum of the squares of any 3, 4, 5 or 6 consecutive integers cant be a perfect square.also give an example of 11 consecutive integers, the sum of whose squares is a perfect square. And Also,Prove that  y^2 - 6 cant be a perfect square (y is an integer)

## 1 Answers SHAIK AASIF AHAMED
askIITians Faculty 74 Points
6 years ago
Hello student,
Please find the answer to your question below
for 3 consecutive numbers:
(n-1)^2+n^2+(n+1)^2
=3n^2+2, which cannot be a perfect square, as it leaves a remainder of 2 when divided by three.
For 4 consecutive numbers:
(n-1)^2+n^2+(n+1)^2+(n+2)^2
=4n^2+4n+6,
which leaves a remainder of 2 upon division by 4.
So even this cannot be a perfect square.
For 5 consecutive numbers:
(n-2)^2+(n-1)^2+n^2+(n+1)^2+(n+2)^2
=5n^2+10
=5(n^2+2)
If this is a perfect square, then
So n^2+2 is in the form 5y^2 for some integer y.
So n^2+2 is divisible by 5.
But n^2 leaves remainder of 0,1 or 4 upon division by 5.
So n^2+2 leaves a remainder of 2,3 or 1 upon division by 5, a contradiction.
So 5(n^2+2) is not a perfect square.
For 6 consecutive numbers:
(n-2)^2+(n-1)^2+n^2+(n+1)^2+(n+2)^2+(n...
=6n^2+6n+19
=6(n^2+n+19)
But n^2+n is always even,
This means that n^2+n+19 is always odd.
So 6(n^2+n+19) is divisible by 2 but not 4.
So it is also not a perfect square.

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