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Bhargava's Theorem

Let the $n$th composition of a function $f(x)$ be denoted $f^{(n)}(x)$, such that $f^{(0)}(x)=x$ and $f^{(1)}(x)=f(x)$. Denote $f\circ g(x)=f(g(x))$, and define

\sum F(a,b,c)=F(a,b,c)+F(b,c,a)+F(c,a,b).
\end{displaymath} (1)

$\displaystyle u$ $\textstyle \equiv$ $\displaystyle (a,b,c)$ (2)
$\displaystyle \vert u\vert$ $\textstyle \equiv$ $\displaystyle a+b+c$ (3)
$\displaystyle \vert\vert u\vert\vert$ $\textstyle \equiv$ $\displaystyle a^4+b^4+c^4,$ (4)

$\displaystyle f(u)$ $\textstyle =$ $\displaystyle (f_1(u),f_2(u),f_3(u))$ (5)
  $\textstyle =$ $\displaystyle (a(b-c), b(c-a), c(a-b))$ (6)
$\displaystyle g(u)$ $\textstyle =$ $\displaystyle (g_1(u),g_2(u),g_3(u))$  
  $\textstyle =$ $\displaystyle \left({\,\sum a^2b, \sum ab^2, 3abc}\right).$ (7)

Then if $\vert u\vert=0$,
$\displaystyle \vert\vert f^{(m)}\circ g^{(n)}(u)\vert\vert$ $\textstyle =$ $\displaystyle 2(ab+bc+ca)^{2^{m+1}3^n}$  
  $\textstyle =$ $\displaystyle \vert\vert g^{(n)}\circ f^{(m)}(u)\vert\vert,$ (8)

where $m,n\in$ $\{0, 1, \ldots\}$ and composition is done in terms of components.

See also Diophantine Equation--Quartic, Ford's Theorem


Berndt, B. C. Ramanujan's Notebooks, Part IV. New York: Springer-Verlag, pp. 97-100, 1994.

Bhargava, S. ``On a Family of Ramanujan's Formulas for Sums of Fourth Powers.'' Ganita 43, 63-67, 1992.

© 1996-9 Eric W. Weisstein