Other solutions for 80abc(a^2+b^2+c^2) = d^5+e^5?

[TPiezas]

Hello all,

There is a basic identity by Lander,

(a+b+c)^5 = (-a+b+c)^5 + (a-b+c)^5 + (a+b-c)^5 + 80abc(a^2+b^2+c^2)

Problem: Find {a,b,c} such that,

80abc(a^2+b^2+c^2) = d^5+e^5 (eq.1)

I found this has an infinite number of non-trivial solns given by,

{a,b,c} = {u^5, 25v^5, 50v^5}, and {d,e} = {10u^3v^2, 50uv^4}.

for arbitrary {u,v}. Questions:

1) Is there any other parametrizations to eq.1?
2) What other rational {a,b,c;d,e}, excluding {1,25,32; 20,40}, that
do not belong to this family? (Enough solns may lead to a pattern.)

- Titus

[alainv...@gmail.com]

Dear Titus,

The equality
(a+b+c)^5 - (-a+b+c)^5 - (a-b+c)^5 - (a+b-c)^5 = 80abc(a^2+b^2+c^2)
is just a particular case of part of form^(2n+1).t^(2p+1).u^(2q+1)
of a polynomial g(s,t,u)=
(g(s,t,u)-g(-s,t,u)-g(s,-t,u)-g(s,t,-u)+g(-s,-t,u)+g(s,-t,-u)+
g(-s,t,-u) - g(-s,-t,-u))/8
On (s+t+u)^3
we've got:
(s+t+u)^3-(-s+t+u)^3-(s-t+u)^3-(s+t-u)^3 = 24stu

Alain

[TPiezas]

On Nov 30, 4:20 am, "alainvergh...@gmail.com"

> Alain- Hide quoted text -
>
> - Show quoted text -

I. Yes, the basic form is called Boutin's Identity which generalizes
the difference of two squares into a sum and difference of 2^(k-1) kth
powers. Thus,

(a+b)^2 - (a-b)^2 = 2ab
(a+b+c)^3 - (a-b+c)^3 - (a+b-c)^3 + (a-b-c)^3 = 24abc

and so on for all kth powers. See Boutin's Identity
http://sites.google.com/site/tpiezas/001.

II. I found other solns to 80abc(a^2+b^2+c^2) = d^5+e^5. The list
{a,b,c}; {d,e} so far is:

{1, 25, 32}; {20,40},
{2, 352, 355}; {-328, 388}
{4, 125, 155}; {-70, 190}
{32, 101, 205}; {-280, 340}

and, other than the known family of solns, there doesn't seem to be
rhyme or reason to these.

- Titus

[Gerry]

> and so on for all kth powers.  See Boutin's Identityhttp://sites.google.com/site/tpiezas/001.

>
> II.  I found other solns to 80abc(a^2+b^2+c^2) = d^5+e^5. The list
> {a,b,c}; {d,e} so far is:
>
> {1, 25, 32};  {20,40},
> {2, 352, 355}; {-328, 388}
> {4, 125, 155}; {-70, 190}
> {32, 101, 205}; {-280, 340}
>
> and, other than the known family of solns, there doesn't seem to be
> rhyme or reason to these.
>

> - Titus- Hide quoted text -

>
> - Show quoted text -

Hi Titus ,

two more solutions :

{a,b,c; d,e} {1,2,125 ; 10, 50}
{a,b,c; d,e} {3,4,240 ; 80,100}

As for another equation for :

80abc(a^2+b^2+c^2) = d^5+e^5 (eq.1)

i found that :

C(a,b,d,e) = c

with:

P(a,b,d,e)=(9*a^2*b^2*(d^5+e^5)+sqrt(3)*sqrt(a^4*b^4*(25600*a^2*b^2*
(a^2+b^2)^3+27*(d^5+e^5)^2)))^(1/3);

C(a,b,d,e)=(5^(1/3)*P(a,b,d,e)-40*6^(1/3)*a^2*b^2*(a^2+ b^2)/P
(a,b,d,e))/(2*30^(2/3)*a*b);

(i don't know how to simplify these)

Regards

Gerry

[TPiezas]

> Gerry- Hide quoted text -

>
> - Show quoted text -

Thanks for the solns, Gerry. Yes, it can be solved as a cubic in
either {a,b,c}. Too bad there's no simple criterion involving the
cubic's discriminant D to determine if it has a rational root. (Unlike
the quadratic where D simply is made a square.)

P.S. I wonder if eq.1, with constraints, can be transformed into an
elliptic curve. After all, it has an infinite number of solns.

- Titus