Re: A Diophantine Equation

Gerry Myerson wrote :
In article <1181687045.703954.199920@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
Robert Israel <israel@xxxxxxxxxxx> wrote:

On Jun 12, 1:02 pm, Maury Barbato <mauriziobarb...@xxxxxxxx> wrote:
Robert Israel wrote:
Maury Barbato <mauriziobarb...@xxxxxxxx> writes:
Hello,
consider the integer solutions of the equation
x^2+y^2=2*(z^2)

Does there exist some solution (x,y,z) such that x=/=y
and x=/=-y?

Infinitely many of them. Try:

x = (a+b)^2 - 2 b^2
y = (a+b)^2 - 2 a^2
z = a^2 + b^2

for distinct positive integers a and b.
--

Oh my God ... how did you find this family of solution?
My poor mind doesn't allow me to make much more than
verifying that this is true ...

You could ask Maple:

isolve(x^2+y^2=2*z^2);

I'm impressed that Maple can do that. In earlier times,
we had to do this kind of thing by hand, something like this:
...
Show that A and B are both Gaussian integers,
...

Hi,

My method looks a little different, as Gaussian integers were unknown
at the time of Fermat, which gave the rules for the sum of two
squares.

If z has no prime factor 4k+1, then 2*z^2 is sum of two squares in just
one way = z^2 + z^2
If z has at least one prime factor 4k+1, then z = m*(a^2 + b^2)
with a != b in at least one way, and using Fibonacci relation

(a^2 + b^2)*(c^2 + d^2) = (a*c +/- b*d)^2 + (a*d -/+ b*c)^2

we can deduce x and y in
m^2 * (1^2 + 1^2)*(a^2 + b^2)*(a^2 + b^2) = x^2 + y^2 :

.. |x| = m*|a^2 - b^2 + 2*a*b|
.. |y| = m*|a^2 - b^2 - 2*a*b|

which are basically the same as those given by Robert Israel.
(at factor m and sign)

Regards.

--
Philippe C., mail : chephip+news@xxxxxxx
site : http://chephip.free.fr/ (recreational mathematics)


.

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