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 >> 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.
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
jacob navia <ja...@nospam.org> writes: > 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
<isr...@math.MyUniversitysInitials.ca> wrote: > jacob navia <ja...@nospam.org> writes: > > 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
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.
In article <32916534.1229157420511.JavaMail.jaka...@nitrogen.mathforum.org>, 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).
On Dec 18, 11:05 am, ru...@vesuvius.math.niu.edu (Dave Rusin) wrote:
> In article <32916534.1229157420511.JavaMail.jaka...@nitrogen.mathforum.org>, > 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).
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".
> 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-a159-a2076fbdf...@t26g2000prh.googlegroups.com>, Chip Eastham <hardm...@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 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
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!
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.
