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I came across this identity $$\left(x^2+7xy-9y^2\right)^3+\left(2x^2-5xy+12y^2\right)^3=\left(2x^2+10y^2\right)^3+\left(x^2-9xy-y^2\right)^3$$

Is there a deeper reason why a formula such as this should exist?

Is there some background where the search for it could be motivated, and that it would come naturally?

Gary
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    From the goddess Namagiri. :) – Deepak Oct 30 '18 at 00:50
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    http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.39.9228&rep=rep1&type=pdf ... see the end of page 9 & page 10 ... Oh & there are lots of other gems too. – Donald Splutterwit Oct 30 '18 at 00:54
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    Regarding the comment from Deepak:... Sriinivasa Ramanujan actually did say that the goddess sent him math as he slept and it is true that he often could not explain his methods of discovery. For more on this extraordinary genius I recommend the memoir about him by G.H.Hardy . – DanielWainfleet Oct 30 '18 at 02:19

2 Answers2

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Sorry for the late reply, I was inactive for about 4 years. Anyway, most of Ramanujan's results are special cases of more general and perhaps deeper phenomena.


I. 3rd powers

For your question, a more general algebraic identity is,

$$A^3+B^3+C^3+D^3 = (a^3+b^3+c^3+d^3)(x^2+wy^2)^3$$

and $(A,B,C,D)$ are quadratic forms,

\begin{align} A &= ax^2-v_1xy+bwy^2\\ B &= bx^2+v_1xy+awy^2\\ C &= cx^2+v_2xy+dwy^2\\ D &= dx^2-v_2xy+cwy^2 \end{align}

where {$v_1,\, v_2,\, w$} = {$c^2-d^2,\; a^2-b^2,\; (a+b)(c+d).$} I gave this identity to Mathworld about 20 years ago.

Thus, given a single example of $a^3+b^3+c^3+d^3 =N$ where $N=0$, a cube, or any number of cubes, then one can generate an infinite more via the LHS. For example, try it with the well-known $3^3+4^3+5^3+(-6)^3 = 0$, permute these variables $(a,b,c,d)$ in any order you wish, and you'll have your own Ramanujan-type identity of sums of cubes.


II. 4th powers

Ramanujan also gave,

$$(2x^2+12xy-6y^2)^4+(2x^2-12xy-6y^2)^4+(4x^2-12y^2)^4+(4x^2+12y^2)^4+(3x^2+9y^2)^4 = (5x^2+15y^2)^4$$

that for $(x,y)=(1,0)$ yields the nice,

$$2^4+2^4+4^4+4^4+3^4 = 5^4$$

The general identity (which I also gave to Mathworld) is,

$$A^4+B^4+C^4 = (a^4+b^4+c^4)(x^2+3y^2)^4$$

and $(A,B,C)$ are also quadratic forms,

\begin{align} A &= ax^2+2(b+c)xy-3ay^2\\ B &= bx^2-2(a+c)xy-3by^2\\ C &= cx^2-2(a-b)xy-3cy^2\\ \end{align}

where $\color{brown}{a+b = c}$. Thus, given a single $a^4+b^4+c^4+d^4+e^4 =f^4$ where $a+b = c$, one can find an infinite more.


P.S. There's actually another cubic identity (as well as a $5$th-deg, but I'm too sleepy. I'll add it soon.) More details can be found at "Generalizing Ramanujan's sum of cubes identity?"

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This is one of many Number Theoretic identities that mathematician Ramanujan discovered. It's hard to say exactly what the "deeper reason" such a formula exists, but it was most likely discovered in search of generalized families of Taxicab numbers. The formula shows a method to write a number as in two distinct ways as a sum of two separate cube numbers. For example: \begin{align*} 1729 &= 1^3 + 12^3 \\ &= 9^3 + 10^3 \\ 87539319 &= 167^3 + 436^3 \\ &= 228^3 + 423^3 \\ &= 255^3 + 414^3. \end{align*} (See if you can find which $x,\: y$, and $z$ you need to reach these solutions!)

As a few commentors have pointed out, Ramanujan claimed to have received many formulas like these from the goddess Namagiri, the Hindu Goddess of creativity. It's an interesting bit of mathematical history that I think you would be delighted to learn. Ramanujan was truly an incredible mind.