Coprime pythagorean quadruples or higher tuples?

1. Jul 18, 2009

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 has a method for generating higher tuples but I can't really follow it...

edit: according to this page 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...

Last edited: Jul 19, 2009
2. Jul 19, 2009

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.

3. Jul 26, 2009

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

4. Jul 27, 2009

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.

5. Jul 29, 2009

ramsey2879

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.

Last edited: Jul 29, 2009