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From: alainverghote on 1 Apr 2010 12:52
Good evening,


Well, it isn't a first april fool question:

when (a^2+b^2)(b^2+c^2)(c^2+d^2)(d^2+a^2)
=
( (a^2+b^2+c^2+d^2)bd )^2 ?

a,b,c,d integer.

Alain
From: Robert Israel on 1 Apr 2010 13:49
"alainverghote(a)gmail.com" <alainverghote(a)gmail.com> writes:

> Good evening,
>
>
> Well, it isn't a first april fool question:
>
> when (a^2+b^2)(b^2+c^2)(c^2+d^2)(d^2+a^2)
> =
> ( (a^2+b^2+c^2+d^2)bd )^2 ?
>
> a,b,c,d integer.
>
> Alain

When ac=(+/-)bd
--
Robert Israel israel(a)math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
From: alainverghote on 2 Apr 2010 05:01
On 1 avr, 19:49, Robert Israel <isr...(a)math.MyUniversitysInitials.ca>
wrote:
> "alainvergh...(a)gmail.com" <alainvergh...(a)gmail.com> writes:
> > Good evening,
>
> > Well, it isn't a first april fool question:
>
> > when    (a^2+b^2)(b^2+c^2)(c^2+d^2)(d^2+a^2)
> >                        =
> >              ( (a^2+b^2+c^2+d^2)bd )^2   ?
>
> > a,b,c,d integer.
>
> > Alain
>
> When ac=(+/-)bd
> --
> Robert Israel              isr...(a)math.MyUniversitysInitials.ca
> Department of Mathematics        http://www.math.ubc.ca/~israel
> University of British Columbia            Vancouver, BC, Canada

Dear Robert,

You didn't us tell the way you get this solution.
I start with the property that all such products
can be written as sums of two squares , two of these sums
simplify into ( (a^2+b^2+c^2+d^2)bd )^2 ,for ac = bd and ac=-bd.

Amicalement,
Alain


From: Robert Israel on 2 Apr 2010 15:28
"alainverghote(a)gmail.com" <alainverghote(a)gmail.com> writes:

> On 1 avr, 19:49, Robert Israel <isr...(a)math.MyUniversitysInitials.ca>
> wrote:
> > "alainvergh...(a)gmail.com" <alainvergh...(a)gmail.com> writes:
> > > Good evening,
> >
> > > Well, it isn't a first april fool question:
> >
> > > when =A0 =A0(a^2+b^2)(b^2+c^2)(c^2+d^2)(d^2+a^2)
> > > =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0=3D
> > > =A0 =A0 =A0 =A0 =A0 =A0 =A0( (a^2+b^2+c^2+d^2)bd )^2 =A0 ?
> >
> > > a,b,c,d integer.
> >
> > > Alain
> >
> > When ac=3D(+/-)bd
> > --
> > Robert Israel =A0 =A0 =A0 =A0 =A0 =A0
> > =A0isr...(a)math.MyUniversitysInitial=
> s.ca
> > Department of Mathematics =A0 =A0 =A0 =A0http://www.math.ubc.ca/~israel
> > University of British Columbia =A0 =A0 =A0 =A0 =A0 =A0Vancouver, BC,
> > Cana=
> da
>
> Dear Robert,
>
> You didn't us tell the way you get this solution.
> I start with the property that all such products
> can be written as sums of two squares , two of these sums
> simplify into ( (a^2+b^2+c^2+d^2)bd )^2 ,for ac =3D bd and ac=3D-bd.
>
> Amicalement,
> Alain

I just factored (with Maple's help)
(a^2+b^2)*(b^2+c^2)*(c^2+d^2)*(d^2+a^2) - ( (a^2+b^2+c^2+d^2)*b*d )^2
= (a*c+b*d)*(a*c-b*d)*(b^4+b^2*d^2+b^2*c^2+a^2*b^2+d^4+c^2*d^2+a^2*d^2+a^2*c^2)
Note that the last factor is nonzero for real a,b,c,d unless
b=d=ac=0.
--
Robert Israel israel(a)math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
From: alainverghote on 3 Apr 2010 03:32
On 2 avr, 21:28, Robert Israel <isr...(a)math.MyUniversitysInitials.ca>
wrote:
> "alainvergh...(a)gmail.com" <alainvergh...(a)gmail.com> writes:
> > On 1 avr, 19:49, Robert Israel <isr...(a)math.MyUniversitysInitials.ca>
> > wrote:
> > > "alainvergh...(a)gmail.com" <alainvergh...(a)gmail.com> writes:
> > > > Good evening,
>
> > > > Well, it isn't a first april fool question:
>
> > > > when =A0 =A0(a^2+b^2)(b^2+c^2)(c^2+d^2)(d^2+a^2)
> > > > =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0=3D
> > > > =A0 =A0 =A0 =A0 =A0 =A0 =A0( (a^2+b^2+c^2+d^2)bd )^2 =A0 ?
>
> > > > a,b,c,d integer.
>
> > > > Alain
>
> > > When ac=3D(+/-)bd
> > > --
> > > Robert Israel =A0 =A0 =A0 =A0 =A0 =A0
> > > =A0isr...(a)math.MyUniversitysInitial=
> > s.ca
> > > Department of Mathematics =A0 =A0 =A0 =A0http://www.math.ubc.ca/~israel
> > > University of British Columbia =A0 =A0 =A0 =A0 =A0 =A0Vancouver, BC,
> > > Cana=
> > da
>
> > Dear Robert,
>
> > You didn't us tell the way you get this solution.
> > I start with the property that all such products
> > can be written as  sums of two squares , two of these sums
> > simplify into ( (a^2+b^2+c^2+d^2)bd )^2 ,for ac =3D bd   and ac=3D-bd.
>
> > Amicalement,
> > Alain
>
> I just factored (with Maple's help)
> (a^2+b^2)*(b^2+c^2)*(c^2+d^2)*(d^2+a^2) - ( (a^2+b^2+c^2+d^2)*b*d )^2
> = (a*c+b*d)*(a*c-b*d)*(b^4+b^2*d^2+b^2*c^2+a^2*b^2+d^4+c^2*d^2+a^2*d^2+a^2*c^­2)
> Note that the last factor is nonzero for real a,b,c,d unless
> b=d=ac=0.
> --
> Robert Israel              isr...(a)math.MyUniversitysInitials.ca
> Department of Mathematics        http://www.math.ubc.ca/~israel
> University of British Columbia            Vancouver, BC, Canada- Masquer le texte des messages précédents -
>
> - Afficher le texte des messages précédents -

Thanks for your answer.
I think dedicated softwares like Maple can be helpful but
not sure about what is learnt out:
(a^2+b^2)*(b^2+c^2)*(c^2+d^2)*(d^2+e^2)*(e^2+a^2) ??

Does it exist any kind of generalization?

Wish you a good Easter day,

Alain
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