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

On Apr 14, 4:12 pm, "Philippe 92" <nos...@xxxxxxxxxxxx> wrote:
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+n...@xxxxxxx
site :http://chephip.free.fr/ ; (recreational mathematics)


Bonjour Philippe,

to find the solutions i'm using the following
equations for every factor q :

p = ( a + b ) / q
x = abs( 2 * p - a )
y = sqrtint( q-2 ) * 2 * p ; (q-2 is a square!)
z = abs(( q-2 ) * p - a )

These are a result of my TWIN^2 identity.

I noticed that i removed the factors 6,66,198 in the examples:
So for N=593^2 + 1^2 = 351650 we get:
The corrected list q=[2, 3, 6, 11, 18, 27, 66, 198]

q=2 N = 1^2 + 0^2 + 593^2
q=3 N = 197^2 + 396^2 + 395^2
q=6 N = 395^2 + 396^2 + 197^2
q=11 N = 485^2 + 324^2 + 107^2
q=18 N = 527^2 + 264^2 + 65^2
q=27 N = 549^2 + 220^2 + 43^2
q=66 N = 575^2 + 144^2 + 17^2
q=198 N = 587^2 + 84^2 + 5^2

Your other 2 square solutions give:

q=2 N = 145^2 + 0^2 + 575^2
q=3 N = 95^2 + 480^2 + 335^2
q=6 N = 335^2 + 480^2 + 95^2
q=18 N = 495^2 + 320^2 + 65^2

q=2 N = 167^2 + 0^2 + 569^2

q=2 N = 229^2 + 0^2 + 547^2

q=2 N = 355^2 + 0^2 + 475^2
q=83 N = 455^2 + 180^2 + 335^2

q=2 N = 373^2 + 0^2 + 461^2
q=3 N = 95^2 + 556^2 + 183^2
q=6 N = 183^2 + 556^2 + 95^2

As you can see for q=3 and q=6 the solutions are the same.
----------

Also your 14 solutions [X,Y,Z] give 8 more solutions
by solving :

N=X^2+Y^2+Z^2=(2*x1+x2)^2 + (2*x1+x3)^2 + (-x1+2*x2-x3)^2

x2 = Y
x3 = 2 * x2 - X
x1 = ( Z - x3 ) / 3

N =465^2 + 368^2 + 1^2 = 188^2 + 91^2 + 555^2
N =161^2 + 552^2 + 145^2 = 20^2 + 411^2 + 427^2
N =345^2 + 460^2 + 145^2
N =575^2 + 143^2 + 24^2
N =575^2 + 87^2 + 116^2
N =575^2 + 17^2 + 144^2
N =575^2 + 105^2 + 100^2 = 415^2 + 55^2 + 420^2
N =231^2 + 520^2 + 167^2 = 92^2 + 381^2 + 445^2
N =547^2 + 221^2 + 60^2 = 331^2 + 5^2 + 492^2
N =133^2 + 456^2 + 355^2
N =285^2 + 380^2 + 355^2 = 300^2 + 395^2 + 325^2
N =475^2 + 213^2 + 284^2 = 435^2 + 173^2 + 364^2
N =261^2 + 380^2 + 373^2 = 296^2 + 415^2 + 303^2
N =461^2 + 275^2 + 252^2
============

Applying the same to your other 2 square
solutions using [X,Y,Z=0] we get 2 solutions
which are already listed:

N =575^2 + 145^2 + 0^2 = 335^2 + 95^2 + 480^2
N =569^2 + 167^2 + 0^2
N =547^2 + 229^2 + 0^2
N =475^2 + 355^2 + 0^2
N =461^2 + 373^2 + 0^2 = 183^2 + 95^2 + 556^2
============
Total solutions found:32

Your score =14 , My score 10+8=18
Your turn ;-)

How did you figure out the total number of solutions?

Do you know an algorithm (not brute force) to find all solutions?

Regards

Gerry
.

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