A Postulate Proof Of A Curious Cubic Identity

Steve McIntyre retweeted a fun tweet yesteryear a statistics professor alongside 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, yous may verify them together with cheque that they work. If yous add together an arbitrary orbit out of the digits \(6,0,3\) to the appropriate places, the identities volition maintain on working. Why is it so? If yous Google search for \(166,500,333\) together with "identity", yous volition speedily observe out that the identity was discussed inwards a 2007 newspaper "A curious cubic identity together with self-similar core of squares" whose PDF file could convey been institute via Google Scholar.

On the peak of page 2/6 (named 70), yous may observe a one-line proof. It looks similar the writer of the proof has been playing alongside diverse curious identities for a century – together with the fact that the newspaper starts alongside a related quote yesteryear Hardy solely strengthens your opinion. The proof is thus fast or uses some impossibly esoteric noesis that it's basically unverifiable unless yous arrive at much to a greater extent than piece of occupation yourself.

But the interplanetary space sequence of identities genuinely has a conceptual argue that yous may observe without knowing anything. Here is the proof.




First, permit me label these identities yesteryear the integer \(k\) which is the orbit out of the digits \(6\) inwards the origin orbit out of the type \(166\). For example, the initial identity adding the cubes of \(1,5,3\) has \(k=0\) together with thus on. Second, to brand both sides converge to a uncomplicated orbit out for \(k\to \infty\), let's dissever the equation yesteryear \(10^{3k}\) to expire an equivalent one.




For example, for \(k=3\), nosotros get\[

0.1666^3+0.5^3+0.3333^3 = 0.1666\,5000\,3333

\] Now it looks simple: nosotros are adding numbers betwixt nil together with one. Moreover, all of them are patch rational. Can nosotros write them downwards every bit a functions of \(k\)? Yes. The origin term on the left manus side is some \((1/6)^3\), isn't it? So it's some equal to \(1/216\). But nosotros tin arrive at better. \(1/6\) would endure equal to \(0.166666\dots\) indefinitely. But if nosotros wish to truncate it, nosotros 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 degree holds for whatever \(k\). Similarly,\[

0.3333 = \frac{1}{3} - \frac{1}{30\times 10^k}

\] because the leading part has doubled together with the modest subtracted slice has halved. The remaining term is just \((1/2)^3=1/8\) together with no complicated subtractions are needed. It's fourth dimension to use\[

(a-b)^3 = a^3 - 3a^2b + 3ab^2 -b^3.

\] We're adding several price together with they volition comprise the factors of \(1,10^{-k},10^{-2k},10^{-3k}\), respectively. Let's grouping them according to the ability of ten. First, the price on the left manus side that comprise no dependence on \(k\) through \(10^{-k}\) etc., i.e. the price resulting from the \(a^3\) price inwards 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 manus side starts alongside \(0.166\dots\). It but had to work. Let's hold off at the correction price alongside the origin ability of \(10^{-k}\). These price arise from \(-3a^2 b\) inwards a formula inwards a higher identify together with they are\[

-\frac{3}{10^k}\times \zav{\frac{1}{6^2\times 15} + \frac{1}{3^2\times xxx } }=\\
-\frac{1}{10^k}\times \zav{\frac{1}{6\times 2\times 15} + \frac{1}{3\times xxx } }

\] In the parentheses, yous watch \(1/180+1/90 = 3/180 = 1/60\) thus when added up, these price are\[

-\frac{1}{60\times 10^{k}},

\] precisely what is needed to trim back \(0.166666\dots\) to \(0.166500\). So this part of the right manus side already works. The solely affair nosotros must expire are the digits \(333\) at the end. But they volition work. Let's expire along alongside the price proportional to \(10^{-2k}\) inwards our expansion, coming from \(3ab^2\) inwards the schoolkid's decomposition above. These price 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 \(10^{-2k} \times (2+1)/900 = 10^{-2k}/300 \), precisely what's needed for \(0.0000003333333\dots\), producing the desired digits \(3\) at the appropriate places on the right manus side. So far, at that spot are infinitely many of them. But the price \(-b^3\) from our schoolkid formula volition 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 reverse sign but otherwise is \(10^{k+1}\) times smaller than the previous term \(10^{-2k}/300\), precisely what's needed to trim back the orbit out of digits \(3\) from infinity to \(k+1\), every bit required. If yous involve to review our simplification of all the price on the left manus side, which indeed matches the right manus side, the seem reads:\[

\frac sixteen - \frac{1}{60\times 10^k} + \frac{1}{300\times 10^{2k}} - \frac{1}{3,000\times 10^{3k}}

\] And that's what nosotros involve to arrive at numbers similar \(0.166650003333\). Such ability constabulary expansions inwards price of the parameter \(1/10^k\) are similar to lots of clever expansions yous may encounter inwards theoretical physics – together with mathematics, of course. Many integers had to piece of occupation nicely. Note that inwards many cases, some excessive factors of \(3\) convey canceled either because of the coefficients \(3\) inwards the expansion of \((a-b)^3\), or because nosotros had things similar \(2+1\) or \(8+1\) inwards the numerators.

I haven't always come across explicit formulae similar to \(36a^3+66a^2+42a+9\) which is mentioned inwards the proof inwards the aforementioned paper. In this sense, my proof looks "inequivalent" although the basic reasons behind the identities are thus straightforward that this inequivalence is just apparent.

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Bonus: As a homework exercise, test a similar interplanetary space sequence of cubic identities such every bit (hat tip: Paul Matthews)\[

3333^3 + 6667^3 + 0001^3 = 3333\,6667\,0001

\]