## Friday, February 03, 2017

### A straight proof of a curious cubic identity

Steve McIntyre retweeted a fun tweet by a statistics professor with three surprising identities:\begin{align} 1^3 + 5^3 + 3^3 &= 153\\ 16^3 + 50^3 + 33^3 &= 16\,50\,33\\ 166^3 + 500^3 + 333^3 &= 166,500,333\\ \end{align} Yes, you may verify them and check that they work. If you add an arbitrary number of the digits $$6,0,3$$ to the appropriate places, the identities will keep on working. Why is it so? If you Google search for $$166,500,333$$ and "identity", you will quickly find out that the identity was discussed in a 2007 paper "A curious cubic identity and self-similar sum of squares" whose PDF file could have been found via Google Scholar.

On the top of page 2/6 (named 70), you may find a one-line proof. It looks like the author of the proof has been playing with various curious identities for a century – and the fact that the paper starts with a related quote by Hardy only strengthens your opinion. The proof is so fast or uses some impossibly esoteric knowledge that it's basically unverifiable unless you do much more work yourself.

But the infinite sequence of identities actually has a conceptual reason that you may find without knowing anything. Here is the proof.

First, let me label these identities by the integer $$k$$ which is the number of the digits $$6$$ in the first number of the type $$166$$. For example, the initial identity adding the cubes of $$1,5,3$$ has $$k=0$$ and so on. Second, to make both sides converge to a simple number for $$k\to \infty$$, let's divide the equation by $$10^{3k}$$ to get an equivalent one.

For example, for $$k=3$$, we get$0.1666^3+0.5^3+0.3333^3 = 0.1666\,5000\,3333$ Now it looks simple: we are adding numbers between zero and one. Moreover, all of them are obviously rational. Can we write them down as a functions of $$k$$? Yes. The first term on the left hand side is approximately $$(1/6)^3$$, isn't it? So it's approximately equal to $$1/216$$. But we can do better. $$1/6$$ would be equal to $$0.166666\dots$$ indefinitely. But if we want to truncate it, we must subtract $$0.0666\dots\times 10^{-k}$$. But $$0.0666\dots = 2/30=1/15$$ so$0.1666 = \frac{1}{6} - \frac{1}{15\times 10^k}$ for $$k=3$$ but this form holds for any $$k$$. Similarly,$0.3333 = \frac{1}{3} - \frac{1}{30\times 10^k}$ because the leading part has doubled and the small subtracted piece has halved. The remaining term is just $$(1/2)^3=1/8$$ and no complicated subtractions are needed. It's time to use$(a-b)^3 = a^3 - 3a^2b + 3ab^2 -b^3.$ We're adding several terms and they will contain the factors of $$1,10^{-k},10^{-2k},10^{-3k}$$, respectively. Let's group them according to the power of ten. First, the terms on the left hand side that contain no dependence on $$k$$ through $$10^{-k}$$ etc., i.e. the terms resulting from the $$a^3$$ terms in the schoolkid's formula above, are$\zav{\frac{1}{6}}^3 \!\!+ \zav{\frac 13}^3 \!\!+ \zav{\frac 12}^3 = \frac{1+8}{2^3\cdot 3^3} + \frac{1}{8} = \frac{1+3}{24}=\frac 16.$ It explains why the right hand side starts with $$0.166\dots$$. It simply had to work. Let's look at the correction terms with the first power of $$10^{-k}$$. These terms arise from $$-3a^2 b$$ in a formula above and they are$-\frac{3}{10^k}\times \zav{\frac{1}{6^2\times 15} + \frac{1}{3^2\times 30 } }=\\ -\frac{1}{10^k}\times \zav{\frac{1}{6\times 2\times 15} + \frac{1}{3\times 30 } }$ In the parentheses, you see $$1/180+1/90 = 3/180 = 1/60$$ so when added up, these terms are$-\frac{1}{60\times 10^{k}},$ exactly what is needed to reduce $$0.166666\dots$$ to $$0.166500$$. So this part of the right hand side already works. The only thing we must get are the digits $$333$$ at the end. But they will work. Let's continue with the terms proportional to $$10^{-2k}$$ in our expansion, coming from $$3ab^2$$ in the schoolkid's decomposition above. These terms are$\frac{3}{10^{2k}} \times \zav{ \frac{1}{6\times 15^2} + \frac{1}{3\times 30^2}} =\\ \frac{1}{10^{2k}} \times \zav{ \frac{1}{2\times 15^2} + \frac{1}{30^2}}$ That's $$10^{-2k} \times (2+1)/900 = 10^{-2k}/300$$, exactly what's needed for $$0.0000003333333\dots$$, producing the desired digits $$3$$ at the appropriate places on the right hand side. So far, there are infinitely many of them. But the terms $$-b^3$$ from our schoolkid formula will truncate those:$-\frac{1}{10^{3k}}\times \zav{ \frac{1}{15^3}+\frac{1}{30^3} } = -\frac{8+1}{10^{3k} \times 27,000 } = \frac{1}{10^{3k}\times 3,000},$ which has the opposite sign but otherwise is $$10^{k+1}$$ times smaller than the previous term $$10^{-2k}/300$$, exactly what's needed to reduce the number of digits $$3$$ from infinity to $$k+1$$, as required. If you need to review our simplification of all the terms on the left hand side, which indeed matches the right hand side, the expression reads:$\frac 16 - \frac{1}{60\times 10^k} + \frac{1}{300\times 10^{2k}} - \frac{1}{3,000\times 10^{3k}}$ And that's what we need to produce numbers like $$0.166650003333$$. Such power law expansions in terms of the parameter $$1/10^k$$ are similar to lots of clever expansions you may encounter in theoretical physics – and mathematics, of course. Many integers had to work nicely. Note that in many cases, some excessive factors of $$3$$ have canceled either because of the coefficients $$3$$ in the expansion of $$(a-b)^3$$, or because we had things like $$2+1$$ or $$8+1$$ in the numerators.

I haven't ever run into explicit formulae similar to $$36a^3+66a^2+42a+9$$ which is mentioned in the proof in the aforementioned paper. In this sense, my proof looks "inequivalent" although the basic reasons behind the identities are so straightforward that this inequivalence is just apparent.

Bonus: As a homework exercise, prove a similar infinite sequence of cubic identities such as (hat tip: Paul Matthews)$3333^3 + 6667^3 + 0001^3 = 3333\,6667\,0001$