Re: --Integer solutions for N=a^2+b^2=x^2+y^2+z^2

Gerry wrote :
On Apr 14, 12:13 am, Robert Israel wrote:
"Philippe 92" writes:
Gerry wrote :

The sum of two squares  N=a^2+b^2
has a solution in three squares  N=x^2+y^2+z^2
for every factor q of the sum a+b
for which q-2 is a square.

Hi Philippe & Robert,

i don't see anything wrong with my statement.

Ok, after your next example, perfectly clear now :
"for every q ..., we get a solution of ..."
or in your exact initial wording :
"has a solution for every q..."
(= at least as many solutions as suitable values of q, including x=0)

Sorry for not understanding at once and taking the blank lines
as meaningfull...

Obviously Philippe spotted the special case
N=20=0^2+2^2+4^2 which indeed has q=6 and q=3
for which q-2 is square.
There are no other representations in this case.
Like Robert points out 0 solutions are not excluded.
Of course these are not of interest.

To clarify for "every factor" consider the following:

Thanks, it helped me understand ;-)

Take a+b = 2*3^3*11 = 594
The list of factors q for which q-2 is a square is:

q=[2, 3, 11, 18, 27]

Now we choose any a < 594 say a=1
Then b = 594-a = 593 and N = 1^2 + 593^2 = 351650
The solutions x^2 + y^2 + z^2 for every q are:
q=2 N = 0^2 + 1^2 + 593^2
q=3 N = 395^2 + 396^2 + 197^2
q=11 N = 107^2 + 324^2 + 485^2
q=18 N = 65^2 + 264^2 + 527^2
q=27 N = 43^2 + 220^2 + 549^2

[I added a few spaces to improve readability]

I didn't find how you deduce (x,y,z) from (a,b,q)
It seems from your examples that q=2 implies x=0, but other q ?

[...]

What i did notice is that 2 can be added
to the list of factors even if a+b is odd.


If true (q=2 => x=0), this just adds 0^2 + a^2 + b^2 to the list ?
Of course N = a^2 + b^2 = 0^2 + a^2 + b^2...

And related with a=2, b=4, q = [3, 6 (and 2) ]
what do you get for q=3 and 6 here ?

So there certainly is a relationship between
the two and three square solutions


Between *some* solutions.

As you missed the 208 other representations of 351650 as sum of
three (nonzero) squares.

My question was are there other relationships like this.


No idea...

I assume in the sequel all squares are nonzero.
Of course if one of a^2 or b^2 is sum of two squares a^2 = u^2 + v^2
(at least one (4k+1) prime factor of a or b), you get u^2 + v^2 + b^2.
I guess you don't mean that.

Example :

N = 351650 = 593^2 + 1^2 = 575^2 + 145^2 = 569^2 + 167^2 =
547^2 + 229^2 = 475^2 + 355^2 = 461^2 + 373^2

593^2 + 1^2,
593^2 = 465^2 + 368^2 gives 465^2 + 368^2 + 1^2

575^2 + 145^2
575^2 = 161^2+552^2 = 345^2+460^2 gives two more solutions
145^2 = 143^2+24^2 = 87^2+116^2 = 17^2+144^2 = 105^2+100^2, 4 more

etc...

So the 6 two squares solutions give 14 three squares solutions :
465^2 + 368^2 + 1^2
161^2 + 552^2 + 145^2
345^2 + 460^2 + 145^2
575^2 + 143^2 + 24^2
575^2 + 87^2 + 116^2
575^2 + 17^2 + 144^2
575^2 + 105^2 + 100^2
231^2 + 520^2 + 167^2
547^2 + 221^2 + 60^2
133^2 + 456^2 + 355^2
285^2 + 380^2 + 355^2
475^2 + 213^2 + 284^2
261^2 + 380^2 + 373^2
461^2 + 275^2 + 252^2

none of them being in your list for 351650 from a = 1, b = 593

Regards.

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


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