Coprime pythagorean quadruples or higher tuples?

[JesseM]

Kind of a random question, but it came up in an online discussion I was having recently about a supposed proof that hinged on pythagorean triples and whether it could be generalized...I know it's possible to find pythagorean triples of the form a^2 + b^2 = c^2 such that a,b,c are all pairwise coprime (no two share a common factor larger than 1), for example 3^2 + 4^2 = 5^2. But is this possible with quadruples of the form a^2 + b^2 + c^2 = d^2, or quintuples of the form a^2 + b^2 + c^2 + d^2 = e^2, or any higher tuples? If so can anyone find some examples? I see this page (http://www.geocities.com/pythagorean_tuples/) has a method for generating higher tuples but I can't really follow it...

edit: according to this page (http://en.wikipedia.org/wiki/Pythagorean_quadruple) the parametrization for generating all primitive pythagorean quadruples apparently implies that at least two of them must be divisible by 2, so in this case it won't work...but I'm still wondering about quintuples and higher...

[JesseM]

It occurred to me that a simple example of a Pythagorean quintuple that's pairwise coprime would just be 1^2 + 1^2 + 1^2 + 1^2 = 2^2. But I wonder if there are any examples where all the numbers are larger than 1.

[JesseM]

Didn't get any response to this one, but so no one wastes their time in the future, just wanted to say that I did find a pairwise coprime pythagorean quintuplet with all numbers greater than 1:

5^2 + 7^2 + 31^2 + 101^2 = 106^2

[robert Ihnot]

This is a little off the subject, but I think there is only one case where consecutive squares starting with 1 is a square:

1^2+2^2 +3^2++++n^2 = \frac{(n)(n+1)(2n+1)}{6} = X^2

And that case is n=24. Obviously they are not all pairwise coprime.

[ramsey2879]

This is a little off the subject, but I think there is only one case where consecutive squares starting with 1 is a square:

1^2+2^2 +3^2++++n^2 = \frac{(n)(n+1)(2n+1)}{6} = X^2

And that case is n=24. Obviously they are not all pairwise coprime.
But it easily would be on topic to add up all the composite squares 16 + 36 + 81 + ... + 576. If that sum is not a coprime square, test if an even number of the prime squares, or if 4 and an even number of odd prime squares, could be added to this sum to make a square and be coprime with the remaining prime squares.