Bhargava's Theorem

Let the th composition of a function 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

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

(1)

Let

$\displaystyle u$	$\textstyle \equiv$	$\displaystyle (a,b,c)$	(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)

and

$\displaystyle f(u)$	$\textstyle =$	$\displaystyle (f_1(u),f_2(u),f_3(u))$	(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.

References

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.