Re: How many rational-not integral points are there on the curve x^2+y^2-2=

In article <20080606.070925@xxxxxxxx>,
I wrote:
In article <6268ad08-e557-41a9-a6c2-2a5497a05f21@xxxxxxxxxxxxxxxxxxxxxxxxxxx>,
Pubkeybreaker <pubkeybreaker@xxxxxxx> wrote:
On Jun 6, 7:00 am, r...@xxxxxxxxxxxxxx (Rob Johnson) wrote:
In article <20080605.215...@xxxxxxxx>,
I wrote:
In article <48489225$0$30461$afc38...@xxxxxxxxxxxxxxxxxxxx>,
"Peter Webb" <webbfam...@xxxxxxxxxxxxxxxxxxxxxxxxx> wrote:
"Gerry Myerson" <ge...@xxxxxxxxxxxxxxxxxxxxxxxxx> wrote in message
news:gerry-81110B.10205006062008@xxxxxxxxxxxxxxxxxxxxx
In article
<31435907.1212708279758.JavaMail.jaka...@xxxxxxxxxxxxxxxxxxxxxx>,
lantan <lantan10101...@xxxxxxxxx> wrote:

Please tell me where the proof is.

The proof of what?

I don't know whether you saw my earlier post in this thread.
Your curve is a circle, and the point (1, 1) is on it.
Draw a line through (1, 1) with rational slope
and find the coordinates of the other point where that line
meets the circle.

Clever, but wrong (I think).

The intersection of the line with the circle occurs at (x,y). Either both of
x and y are rational, or both are irrational. That's as much as you can say.

I suspect the answer is exactly the set of pythagorean triples, as all such
solutions if they are rational can be multiplied by the lcm of the
denominators to provide a pythagorean triple (Johnson proved that it is at
least the Pythagorean triples, but it seems obvious that this is also all
solutions).

Indeed. Suppose x+iy is a rational point on the circle of radius 2.
Then |x+iy|/|1+i| = 1. Therefore,

x+iy (x+iy)(1-i) (x+y) + i(y-x)
---- = ----------- = --------------
1+i 2 2

is a rational point on the circle of radius 1, that is, a point
derived from a Pythagorean Triple. Thus,

(x+y) + i(y-x)
x+iy = -------------- (1+i)
2

Follow up question: how many rational points are on the circle of
radius 3?

Instead of "circle of radius 2" and "circle of radius 3", I should
have said "circle of radius sqrt(2)" and "circle of radius sqrt(3)".

In particular, the followup question should be: how many rational
points are on the circle of radius sqrt(3)?

Rob Johnson <r...@xxxxxxxxxxxxxx>

Where is the center??? Assuming it is the origin, the answer is none.
Integers that are 3 mod 4 are not the sum of two squares.

I should have been more explicit, but I was assuming that we were
talking about curves such as that mentioned in the title, where the
center is the origin.

You are correct that no integer that is 3 mod 4 is the sum of two
squares. However, this does not answer the question of whether there
are any rational pairs (x,y) so that x^2 + y^2 = 3. This is related
to the diophantine equation x^2 + y^2 = 3 z^2.

Actually, you can build a proof around 3 mod 4 not being the sum of
two squares (i.e. 3 z^2 is either 0 or 3 mod 4, and 0 mod 4 is not
the sum of two relatively prime squares). I was thinking of a proof
based on the fact that 0 mod 3 is not the sum of two relatively prime
squares. The second is a little more direct, but no more correct.

Rob Johnson <rob@xxxxxxxxxxxxxx>
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