Let $$\small F\_n=(a+b+c)^n+(b+c+d)^n-(c+d+a)^n-(d+a+b)^n+(a-d)^n-(b-c)^n$$ and
$ad=bc$, then
$$64 F\_6 F\_{10}=45 F\_8^2$$
This fascinating identity is due to Ramanujan and can be found in ["Ramanujan for Lowbrows", by B.C. Berndt and S. Bhargava](http://www.maa.org/programs/maa-awards/writing-awards/ramanujan-for-lowbrows). Would anyone have any idea how Ramanujan discovered this identity?


The proofs of the identity offered so far in the papers ["A Note on an Identity of Ramanujan", by T. S. Nanjundiah](http://www.jstor.org/discover/2324305), ["Two or Three Identities of Ramanujan", by M.D. Hirschhorn](http://www.jstor.org/discover/10.2307/2589526) and ["A remarkable identity found in Ramanujan's third notebook", by B.C. Berndt and S. Bhargava](http://journals.cambridge.org/article_S0017089500008910) make the identity less mysterious, but how Ramanujan found the identity in the first place still remains a mystery. As Berndt and Bhargava remarked, it is also not clear whether this is an accidental, isolated result (along with the 3-7-5 counterpart discovered by Hirschhorn), or if there is some deeper theorem lurking behind it.



How did Ramanujan discover this identity?
You have two questions: 1) How Ramanujan discovered it? 2) Is an accidental, isolated result? The second one is easier to answer and may shed light on the first.



> 
> **I.** Define $F\_n = x\_1^n+x\_2^n+x\_3^n-(y\_1^n+y\_2^n+y\_3^n),\;$ where $\,\small x\_1+x\_2+x\_3=y\_1+y\_2+y\_3 = 0$.
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**Theorem 1:** If $F\_1 = F\_3 = 0$, then,


$$9x\_1x\_2x\_3 F\_6 = 2F\_9 = 9y\_1y\_2y\_3 F\_6$$


**Theorem 2:** If $F\_2 = F\_4 = 0$, then,


$$64F\_6F\_{10} = 45F\_8^2\quad \text{(Ramanujan)}$$


$$25F\_3F\_{7} = 21F\_5^2\quad \text{(Hirschhorn)}$$



> 
> **II.** Define $F\_n = x\_1^n+x\_2^n+x\_3^n+x\_4^n-(y\_1^n+y\_2^n+y\_3^n+y\_4^n),\;$ where also $\,\small \sum x\_i =\sum y\_i= 0$.
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**Theorem 3:** If $F\_1 = F\_3 = F\_5 = 0$, then,


$$7F\_4F\_9 = 12F\_6F\_7\quad \text{(yours truly)}$$


This also has a similar Ramanujan-type formulation. Define,


$$\small P\_n = ((a+b+c)^n + (a-b-c)^n + (-a-b+c)^n + (-a+b-c)^n – ((d+e+f)^n + (d-e-f)^n + (-d-e+f)^n + (-d+e-f)^n)$$


If *two* conditions are satisfied,


$$\small abc = def,\quad a^2+b^2+c^2 = d^2+e^2+f^2$$


then,


$$7P\_4P\_9 = 12P\_6P\_7$$


(The two conditions have an infinite number of primitive solutions, one of which is $1,10,12;\,2,4,15$.) 


If you are looking for the general theory behind Ramanujan's 6-10-8 Identity, the theorems flow from the properties of ***equal sums of like powers***. The 6-10-8 needed only ***one*** condition, namely $ad=bc$. Going higher, you now need ***two***. Presumably going even higher would need more. Also, there is a constraint $\,\small \sum x\_i =\sum y\_i= 0$. Without this constraint, then more generally,



> 
> **III.** Define $F\_n = x\_1^n+x\_2^n+x\_3^n-(y\_1^n+y\_2^n+y\_3^n),\;$ and $m = (\sum x\_i^4)/(\sum x\_i^2)^2$.
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**Theorem 4:** If $F\_2 = F\_4 = 0$, then,


$$32F\_6F\_{10} = 15(m+1)F\_8^2$$


(*Note:* Ramanujan's simply was the case $m=1/2$.)



> 
> **IV.** Define $F\_n = x\_1^n+x\_2^n+x\_3^n+x\_4^n-(y\_1^n+y\_2^n+y\_3^n+y\_4^n),\;$ and $m = (\sum x\_i^4)/(\sum x\_i^2)^2$.
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**Theorem 5:** If $F\_2 = F\_4 = F\_6 = 0$, then,


$$25F\_8F\_{12} = 12(m+1)F\_{10}^2$$


and so on for similar identities with more terms and *multi-grade* higher powers, *ad infinitum*.


**V. Conclusion:** Thus, was the 6-10-8 Identity an isolated result? No, it is a special case (and a particularly beautiful one at that) of a more general phenomenon. And how did Ramanujan find it? Like most of his discoveries, he plucked it out of thin air, I suppose.


**P.S.** Humor aside, what I read was that, as he found paper expensive, he would scribble on a small slateboard with chalk. After he was satisfied with a result, he would write it down on his notebook, and *erase the intermediate steps that was on the slateboard*. (Sigh.)


Also, since he spent most of his waking hours thinking about mathematics, I think it was only natural it spilled over into his dream state. (A similar thing happened with the chemist Kekule and the discovery of the benzene ring.)


**P.P.S.** Results for multi-grades can be found in Table 2 of this [MO answer](https://mathoverflow.net/a/191939/12905). 


