9
$\begingroup$

Let $ad=bc$. Then Ramanujan's 6-10-8 Identity is the bizarre,

$$\small 64[(a+b+c)^6+(b+c+d)^6-(c+d+a)^6-(d+a+b)^6+(a-d)^6-(b-c)^6]\\ \small [(a+b+c)^{10}+(b+c+d)^{10}-(c+d+a)^{10}-(d+a+b)^{10}+(a-d)^{10}-(b-c)^{10}] =45[(a+b+c)^8+(b+c+d)^8-(c+d+a)^8-(d+a+b)^8+(a-d)^8-(b-c)^8]^2$$

It can be somewhat de-mystified. For generic $a,b,c,d,e,f$, define,

$$F_k = a^k+b^k+c^k-(d^k+e^k+f^k)$$

If $F_2 = F_4 = 0$ and $a+b+c = d+e+f = 0$, then the 6-10-8 is,

$$64F_6 F_{10} = 45F_8^2$$

and Hirschhorn gave,

$$25F_3 F_7 = 21F_5^2$$

I found this has an odd power counterpart. Define,

$$G_k = a^k+b^k+c^k+d^k-(e^k+f^k+g^k+h^k)$$

If $G_1 = G_3 = G_5 = 0$ and $a+b+c+d = e+f+g+h = 0$, then it can be observed that,

$$7G_4G_9 = 12 G_6G_7$$

One can test it with a random example {$a,b,c,d$} = {$21,\, 9, -13, -17$} and {$e,f,g,h$} = {$23,\, 1, -3, -21$} but it will work in general.

Question: The next step seems obvious: to use more terms and higher powers. Anybody knows how to find if there are higher power versions?

$\endgroup$
  • 2
    $\begingroup$ No idea. I know how to search for such things, though, by giving very small absolute values for all variables, then see if your $H_r H_s$ and $H_t H_{r+s-t}$ keep a constant ratio. Save failures in one file. Save candidates to another file. Run candidates with somewhat larger absolute values. Run survivors pretty high, finally do completely in symbols in pari or the like. $\endgroup$ – Will Jagy May 6 '13 at 5:11
  • $\begingroup$ An identity using the $15$th power, $957P_8 P_{15} =1547P_{10} P_{13}$ is possible. See this MO post. $\endgroup$ – Tito Piezas III Mar 4 '16 at 22:35
1
$\begingroup$

I found simplified equivalent expressions by changing $F_k$ to $P_k$ $($where $P_k=F_k/k)$.

Define $$P_k=(a_1^k+a_2^k+a_3^k-b_1^k-b_2^k-b_3^k)/k$$ If $P_1=P_2=P_4=0$, then the Ramanujan's 6-10-8 Identity is, $$4P_6P_{10}=3P_8^2$$ and the Hirschhorn's 3-7-5 Identity is, $$P_3P_7=P_5^2$$ and the identities by Chamberland are, $$P_3P_8=2P_5P_6$$ $$P_3P_{10}=3P_6P_7$$

I found more interesting results, $$\frac{P_5}{P_3}=\frac{P_7}{P_5}=\frac{P_8}{2P_6}=\frac{2P_{10}}{3P_8}=\sqrt{\frac{P_7}{P_3}}=\sqrt{\frac{P_{10}}{3P_6}}$$ $$\frac{P_8}{2P_5}=\frac{P_6}{P_3}=\frac{P_{10}}{3P_7}$$ $$\frac{3P_5^2}{P_{10}}=\frac{P_3^2}{P_6}=\frac{-P_{-1}^2}{P_{-2}}$$

For higher power, define, $$Q_k=(a_1^k+a_2^k+a_3^k+a_4^k-b_1^k-b_2^k-b_3^k-b_4^k)/k$$ If $Q_1=Q_2=Q_3=Q_5=0$, then the 4-9-6-7 identity can be simplified as, $$Q_4Q_9=2Q_6Q_7$$

I found the next higher power version. Define, $$R_k=(a_1^k+a_2^k+a_3^k+a_4^k+a_5^k-b_1^k-b_2^k-b_3^k-b_4^k-b_5^k)/k$$ If $R_1=R_2=R_3=R_4=R_6=0$, then, $$R_5(R_8^2+2R_7R_9-R_5R_{11})=R_7^3$$ Example $$a_1=-23,\ a_2=-10,\ a_3=-5,\ a_4=14,\ a_5=24$$ $$b_1=-21,\ b_2=-16,\ b_3=2,\ b_4=10,\ b_5=25$$ For more numeral examples, see my site on Equal Sums of Like Powers.

For more similar identities, see my site on The Generalization of Ramanujan’s 6-10-8 identity

or refer to the following pictures.

Simplified equivalent expressions of Ramanujan’s identity found by Chen Shuwen

Higher power version of Ramanujan’s 6-10-8 identity found and proved by Chen Shuwen

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.