Diophantine Equation x^2 = a^2 + b^2 + c^2
M. A. Fajjal
Hi
I have noticed that the square of any integer can be represented as the sum of 3 squars except 2^n and 5*2^n, n=0,1,2,...
i.e for any x <> {2^n, 5*2^n} , there exist a ,b ,c such that
x^2 = a^2 + b^2 + c^2
Is there any proof
mma...@personal.ro
Your conjecture fails for x = 24576.
Mate
jacob navia
24576 = 3*(2^12) = 3*8192
Maybe we should add N*(2^n) N integer 1,2,3... to the exceptions list?
--
jacob navia
jacob at jacob point remcomp point fr
logiciels/informatique
http://www.cs.virginia.edu/~lcc-win32
M. A. Fajjal
8192^2+ 16384^2 + 16384^2 = 24576^2
jacob navia
OK, I thought a, b, and c should be all diffferent.
M. A. Fajjal
x^2 = a^2 + b^2 + c^2
x = (p^2+q^2+r^2+s^2)
a = (p^2+q^2-r^2-s^2)
b = (2*p*r+2*q*s)
c = (2*p*s-2*q*r)
Where a > 0, b > 0, c > 0 or
According to Lagrange's Four-Square Theorem which was not proven by Diophantus, every positive integer can be written as the sum of at most four squares
http://mathworld.wolfram.com/LagrangesFour-SquareTheorem.html
For x = {2^n, 5*2^n}
a = 0
or b = 0
or c = 0
Robert Israel
> M. A. Fajjal wrote:
> >> On Dec 13, 10:36 am, "M. A. Fajjal" <H2...@se.com.sa>
> >> wrote:
> >>> Diophantine Equation x^2 = a^2 + b^2 + c^2
> >>>
> >>> Hi
> >>>
> >>> I have noticed that the square of any integer can
> >> be represented as the sum of 3 squars except 2^n and
> >> 5*2^n, n=0,1,2,...
> >>> i.e for any x <> {2^n, 5*2^n} , there exist a ,b ,c
Nonzero, I suppose, else you could always take a,b,c = x,0,0, or
for x = 5*2^n take a=3*2^n, b=4*2^n, c=0.
> >> such that
> >>> x^2 = a^2 + b^2 + c^2
> >>>
> >>> Is there any proof
> >> Your conjecture fails for x = 24576.
> >>
> >> Mate
> >
> > 8192^2+ 16384^2 + 16384^2 = 24576^2
>
> OK, I thought a, b, and c should be all diffferent.
If x is even, x^2 is divisible by 4. Now if a^2 + b^2 + c^2 == 0 mod 4
a, b, c must all be even. Thus the solutions of x^2 = a^2 + b^2 + c^2
with x even correspond to solutions of (x/2)^2 = d^2 + e^2 + f^2
with a=2d, b=2e, f=2c. Iterating, we find that if x is divisible by
2^k, any solution must have a,b,c all divisible by 2^k. In particular,
if x = 2^k or 5*2^k, there is no solution since there are no solutions
for x=1 or x=5. For x=3*2^k, the only solutions will be
x^2 = (2^k)^2 + (2*2^k)^2 + (2*2^k)^2, since the only solution to
3^2 = a^2 + b^2 + c^2 is 3^2 = 1^2 + 2^2 + 2^2.
--
Robert Israel isr...@math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
Chip Eastham
Note that if x is such that x^2 = a^2 + b^2 + c^2
for some integers a,b,c > 0, any integer multiple
of x also has this property. A minimal counter-
example would thus have to be a prime x > 5.
--c
Dave Rusin
M. A. Fajjal <H2...@se.com.sa> wrote:
>I have noticed that the square of any integer can be represented as the sum of 3 squars except 2^n and 5*2^n, n=0,1,2,...
Every positive integer can be expressed as a sum of three squares except
those of the form 4^n ( 8m-1 ) (which clearly cannot be so expressed,
because of a 2-adic condition).
Chip Eastham
> In article <32916534.1229157420511.JavaMail.jaka...@nitrogen.mathforum.org>,
Good to see you around, Dave!
Quite so, however the OP appears to impose a further
condition that the summands should be nonzero, i.e.
the sum of three squares and no less. Otherwise x^2
would evidently be a "sum of three squares".
regards, chip
Dave Rusin
> I have noticed that the square of any integer can be represented as the
> sum of 3 squars except 2^n and 5*2^n, n=0,1,2,...
I wrote,
> Every positive integer can be expressed as a sum of three squares except
> those of the form 4^n ( 8m-1 ) (which clearly cannot be so expressed,
> because of a 2-adic condition).
In article <3e21fd6d-2c23-4dec...@t26g2000prh.googlegroups.com>,
Chip Eastham <hard...@gmail.com> wrote:
> Quite so, however the OP appears to impose a further
> condition that the summands should be nonzero, i.e.
> the sum of three squares and no less. Otherwise x^2
> would evidently be a "sum of three squares".
Right, sorry, I wasn't paying enough attention.
The proposed result is classical. I give you a USENET copy of a Math Reviews
summary of a Monthly article reviewing a textbook reference of a classical
presentation of a clever identity found by Lebesgue:
MR0258742 (41 #3388)
Fraser, Owen; Gordon, Basil
On representing a square as the sum of three squares.
Amer. Math. Monthly 76 1969 922--923.
The authors give an elementary proof of the following theorem stated
by A. Hurwitz [Mathematische Werke. Band 2. Zalentheorie, Algebra und
Geometrie, p. 751, Birkhauser, Basel, 1933; reprinting, 1963;
MR0154778 (27 #4723b)]: The equation $n^2=x^2+y^2+z^2$ has a solution
in positive integers $x,y,z$ if and only if $n$ is not of the form
$2^k$ or $2^k . 5$.
Reviewed by T. Hayashida; Copyright 2008, American Mathematical Society
The AMM article is available through JSTOR and (probably illegally) at
http://www.dm.unito.it/~cerruti/ntlab2007/3squares-elementary.pdf
The authors give a proof of our OP's result, which was stated without proof
by Hurwitz, but I like their opening line, which quotes a book by Nagell
that shows in spirit why the result is true: simply use Lebesgue's identity
(a^2+b^2+c^2+d^2)^2 = (a^2+b^2-c^2-d^2)^2 + (2ac+2bd)^2 + (2ad-2bc)^2
together with the four-square theorem to observe that every square is a
sum of three squares! But as Chip pointed out, this doesn't completely
resolve the issue about making the three terms on the right nonzero.
Too bad, because it's an elegant proof of something or other!
I should also note that someone once asked about writing _cubes_ as sums of
three squares; there are some moderately clever arguments about that here:
http://www.math.niu.edu/~rusin/known-math/96/3squares
There are just oodles of results concerning the representations of numbers
as sums of squares, e.g. linking the number of such representations to
theta functions and so on. It's heavy-duty stuff, but cool.
dave