Moessner's Fifth Powers Identity

[TPiezas]

Hello all,

Moessner (1951) gave the beautiful identity,

(a+20t^11)^5 + (b+20t^25)^5 + (t^33-t^3+c)^5 = (a-20t^11)^5 +
(b-20t^25)^5 + (t^33-t^3+d)^5

where,

a = (t^30+8t^20+12t^10-1)*t^6
b = t^30-12t^20-8t^10-1
c = -12t^23-28t^13
d = 28t^23+12t^13

Notice how the coefficients, after negation, are palindromic. Other
than the even simpler Sastry-Chowla Identity, which _surprisingly_
depends on the Pythagorean triple {3,4,5}, does anybody know of
similar simple 5th power identities?

See more at: http://sites.google.com/site/tpiezas/020

- Titus

[alainv...@gmail.com]

Bonjour Titus,

I.e if we want to start with the equality

(a+20t^11)^5 + (b+20t^25)^5 + (t^33-t^3+c)^5 = (a-20t^11)^5 +
(b-20t^25)^5 + (t^33-t^3+d)^5

we rewrite it:
2*evenpart/[a,b or c]{(a+20t^11)^5+(b+20t^25)^5+(c+t^33-t^3)^5)=0} ,

2*evenpart/[a]{(a+20t^11)^5}
= 5*a^4*20t^11+10*a^2*(20t^11)^3+(20t^11)^5
= 5*a^4*20t^11+10*20^3*a^2*t^33+20^5*t^55
2*evenpart/[b]{(b+20t^25)^5}
= 5*b^4*20t^25+10*b^2*(20t^25)^3+(20t^25)^5
=
2*evenpart/[c]{(c+t^33-t^3)^5)}
= 5*c^4*(t^33-t^3)+10*c^2*(t^33-t^3)^3+(t^33-t^3)^5
=
d= -c ,we must adjust polynomials a(t),b(t),c(t)
to obtain: 2*evenpart/[a]+2*evenpart/[b]+2*evenpart/[c]=0

Hope it helps,

Alain