Re: Does x^2+y^2=z^50 have integer solutions?

On Fri, 09 Jan 2009 04:03:41 +0100, Timothy Murphy
<gayleard@xxxxxxxxxx> wrote:

Bill Dubuque wrote:

Timothy Murphy <gayleard@xxxxxxxxxx> wrote:
alvarez.torres1@xxxxxxxxx wrote:

Does x^2 + y^2 = z^50 have integer solutions >0?

A positive integer n is expressible in the form n = x^2 + y^2
if and only if each prime q = 3 mod 4 occurs to an even power in n.
(See eg Hardy & Wright, "An introduction to the theory of numbers".)

It follows that your equation has a solution for _every_ z.

That's trivially true: 0^2 + (z^25)^2 = z^50, so there's
no need to infer it from theorem in the prior paragraph.

(Well, you need to exclude some z if you want both x,y > 0.)

Since the problem requires solutions >0, your remark doesn't help.

I wasn't purporting to give a complete answer to the question.
I was just pointing out that it reduces to the question
of which n are expressible in the form x^2 + y^2.
I mentioned Hardy & Wright,
reference to which would answer the question completely.

If you want a complete answer, n is expressible as x^2 + y^2
with x,y > 0 iff each prime q = 3 mod 4 occurs to an even power,
and some prime not = 3 mod 4 occurs in n.

So the problem is soluble with any non-zero z
divisible by some prime p not = 3 mod 4.

By some _odd_ prime not = 3 mod 4 (i.e. p = 1 mod 4).

But it looks like that clinches the question I was trying to answer.

Thus, based on your observations, I think the following iff condition
does the trick ...

Proposition:

Let n be a positive integer.

A positive integer z is such that the equation

x^2 + y^2 = z^(2n)

has a solution in positive integers x,y iff z has a prime factor
congruent to 1 mod 4.

quasi
.

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