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. 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, "Two or Three Identities of Ramanujan", by M.D. Hirschhorn and "A remarkable identity found in Ramanujan's third notebook", by B.C. Berndt and S. Bhargava 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.

  • 6
    $\begingroup$ What's the star operator? $\endgroup$
    – Dan Piponi
    Jan 13, 2015 at 18:00
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    $\begingroup$ $*$ is just multiplication. $\endgroup$
    – Stopple
    Jan 13, 2015 at 18:12

5 Answers 5


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 $\,\small x_1+x_2+x_3=y_1+y_2+y_3 = 0$.

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 $\,\small \sum x_i =\sum y_i= 0$.

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,

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


$$7P_4P_9 = 12P_6P_7$$

(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 $m = (\sum x_i^4)/(\sum x_i^2)^2$.

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 $m = (\sum x_i^4)/(\sum x_i^2)^2$.

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.

  • $\begingroup$ This is nice. But in III and IV, what about the $y_i$ in the definition of $m$? Should it be a condition $(\sum x_i^4)/(\sum x_i^2)^2= (\sum y_i^4)/(\sum y_i^2)^2=:m $ ? $\endgroup$
    – Wolfgang
    Jan 17, 2015 at 15:43
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    $\begingroup$ @Wolfgang: No need to define $m$ in terms of $y_i$. If you look at the theorems below them, it assumes $\sum x_i^n = \sum y_i^n$ for $n=2,4$ from which your equality necessarily follows. (For IV, it is $n=2,4,6$.) $\endgroup$ Jan 17, 2015 at 16:55
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    $\begingroup$ @Tito Very nice! Have you published these results? $\endgroup$ Jan 18, 2015 at 7:32
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    $\begingroup$ @ZurabSilagadze: Thank you. No, I haven't yet. It is mostly to satisfy my curiosity as I got intrigued by the 6-10-8 identity when I came across it. $\endgroup$ Jan 18, 2015 at 8:08
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    $\begingroup$ Note that some of these identies are of the form $n^2F_{n-2}F_{n+2}=t_{n+1}F_{n}^{2}$ with $t_m$ the $m$-th triangular number. $\endgroup$ Jul 27, 2019 at 15:51

One line of argument (section II), from a fellow MO-user, starts from the premise that Ramanujan knew that $F_n=0$ for $n=2$ or $n=4$:

It’s almost a shame to give away the secret of the 6-10-8 Identity. To quote the poet Keats, it’s like unweaving a rainbow: things will be different once we know something too well.

  • $\begingroup$ Nice article! However, as it states in Conclusions, "the question remains how he found it in the first place". $\endgroup$ Jan 14, 2015 at 5:01
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    $\begingroup$ One possibility is that the identity was communicated to Ramanujan by the Goddess of Namagiri ): . Ramanujan by himself describes one of such dreams (from the book "Tibetan Dream Yoga", by Michael Katz): "While asleep, I had an unusual experience. There was a red screen formed by flowing blood, as it were. I was observing it. Suddenly a hand began to write on the screen. I became all attention. That hand wrote a number of elliptic integrals. They stuck to my mind. As soon as I woke up, I committed them to writing". $\endgroup$ Jan 14, 2015 at 5:17
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    $\begingroup$ @ZurabSilagadze Yes, that's the obvious answer to how Ramanujan got the identities. But how did the Goddess of Namagiri discover them? $\endgroup$
    – bof
    Jan 16, 2015 at 17:04
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    $\begingroup$ She did it by plugging in values for the variables. That, or she consulted an oracle who performed the infinitely many computations. $\endgroup$ Jan 16, 2015 at 17:37
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    $\begingroup$ @Carlo: I never thought a sentence I wrote ten years ago would find its way here. :) $\endgroup$ Jan 16, 2015 at 20:17

(This is an addendum to my answer.)

I. Identities. For those curious on further generalizations of Ramanujan's 6-10-8, while Theorem 3 used $2^3$ terms, one with higher powers uses $2^5$ terms. Define,

$$\small P_n = \sum^{16}\, (\mu(a_1\pm a_2\pm a_3\pm a_4\pm a_5))^n - \sum^{16}\, (\mu(b_1\pm b_2\pm b_3\pm b_4\pm b_5))^n\tag1$$

where $\mu = \pm1$ and is the product of the interior signs. If four conditions are now satisfied,

$$\small\prod^5 a_i = \prod^5 b_i$$

$$\small \sum^5 a_i^k = \sum^5 b_i^k$$

for $k = 2,4,6$, then,

$$P_1 = P_2 = P_3 = P_4 = P_5 = P_6 = P_7 = P_9 = P_{11} = 0$$


$$957\,P_8 P_{15} = 1547\,P_{10} P_{13}$$

II. Families. I found one can solve the four conditions in two ways: the first in terms of quadratic forms, and the second as an elliptic curve.

Family 1: If $x^2+21y^2 = z^2$, then,

$$a_i = x + 6 y,\; x + 5 y - z,\; x + 5 y + z,\; \tfrac{1}{2}(-5x+2y),\; \tfrac{3}{2}(x-6y)$$

$$b_i = x - 6 y,\; -x + 5 y - z,\; -x + 5 y + z,\; \tfrac{1}{2}(5x+2y),\; \tfrac{3}{2}(x+6y)$$

Family 2: If $-4a^2+5b^2=4c^2,\;\; 25a^2+24b^2=d^2$, then,

$$a_i = 4 a - 4 b,\; 3 b - 2 c,\; 3 b + 2 c,\; 6 a,\; 4 a + 4 b$$

$$b_i = b - 2 c,\; b + 2 c,\; 4 a,\; -a + d,\; a + d$$

Note: For the second family, eight terms will cancel out in $(1)$, so it really involves only $2^5-8=24$ terms. An initial solution is $a,b,c,d = 29, 26, 2, 193$ and, using an elliptic curve, one can get an infinite more. Explicitly,

$$\small( 281, -207, -199, -107, \color{blue}{-61}, 125, \color{blue}{33}, -13, \color{blue}{25}, -21, \color{blue}{-113}, 49, 95, 187, 195, -269)^n=\\ \small(277, -247, -161, \color{blue}{-113}, -55, 131, 83, \color{blue}{25}, -3, \color{blue}{-61}, -109, \color{blue}{33}, 91, 139, 225, -255)^n$$

for $n = 1,2,3,4,5,7,9,11$. (The second family was found in 2013 with the help of Roger Glendenning who provided numerical solutions. After some trial and error, I found the form above.)


How Ramanujan found the identity is, of course, a mystery, but with the intelligent use of a computer algebra system, many such identities can be found. I wrote an article about this. Two of the simpler identities that were not listed above include $$ F_{-2}F_3^2 = -3 F_{-1}^2 F_6 $$ and $$ 245F_3 F_{11} + 330F_7^2 = 539F_5 F_9 $$


I found that Theorem 4 and Theorem 5 can be simplified as below,

Theorem 4(s):

Define $R_n = (x_1^n+x_2^n+x_3^n-y_1^n-y_2^n-y_3^n)/n,\;$ and $m = (\sum x_i^2)/(\sum x_i)^2$ .

If $R_1=R_2=0$, then $$2R_3R_5 = (m+1)R_4^2$$

Theorem 5(s):

Define $R_n = (x_1^n+x_2^n+x_3^n+x_4^n-y_1^n-y_2^n-y_3^n-y_4^n)/n,\;$ and $m = (\sum x_i^2)/(\sum x_i)^2$ .

If $R_1=R_2=R_3=0$, then $$2R_4R_6 = (m+1)R_5^2$$

Futhermore, we have

Define $R_n = (x_1^n+x_2^n+x_3^n+x_4^n+x_5^n-y_1^n-y_2^n-y_3^n-y_4^n-y_5^n)/n,\;$ and $m = (\sum x_i^2)/(\sum x_i)^2$ .

If $R_1=R_2=R_3=R_4=0$, then $$2R_5R_7 = (m+1)R_6^2$$ and so on.

More similar identities can be found on my website on Algebraic Identities.


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