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In the bottom of one of the pages of Ramanujan's notebooks (see below), he writes down the following equations: $$\left(11\frac 12\right)^3+\left(\frac 12\right)^3=39^2$$ $$\left(3\frac 17\right)^3-\left(\frac 17\right)^3=\left(5\frac 47\right)^2$$ $$\left(3-\frac 1{105}\right)^3+\left(\frac 1{105}\right)^3=\left(5\frac 6{35}\right)^2$$ $$\left(3\frac 1{104}\right)^3-\left(\frac 1{104}\right)^3=\left(5\frac{23}{104}\right)^2$$ It is clear he was studying the equation $A^3+B^3=C^2$ just like how Euler did, but he also took it one step further and extended the solutions to the set of rational numbers. Now Ramanujan was particularly fond of finding remarkable general identities, but often his most curious ones was he going to remember the most. So he never really wrote them down, and predominantly wrote special cases of them. From the elegant form of the equations above, I presume he had a general identity.

Empirically I have found a near solution (my best one). $$\left(14+\frac 13\right)^3+\left(\frac 13\right)^3\simeq \left(54+\frac{442}{1665}\right)^2$$

And some solutions with irrationals. $$\left(6-\frac 4{3\pm \sqrt 5}\right)^3+\left(\frac 4{3\pm \sqrt 5}\right)^3=12^2$$

$$\left(1\pm\frac 2{\sqrt 5-1}\right)^3\mp\left(\frac 2{\sqrt 5-1}\right)^3=2^2$$

Anybody know how Ramanujan got these results? I just factored out $(a+b)^3-b^3$ since the $b^3$ terms cancelled out, leaving me with quadratics, which is fairly simple. I think if Ramanujan had a general identity for this, it would do the same. Note that $(-1\frac 12)^3+(\frac 12)^3=0^2$ which is another solution to his first equation, and both solutions can be written as $\frac{10\pm 12}{2}$ and $\frac{39\pm 39}{2}$ which looks quadratic.

Any thoughts?

Thanks.

Ramanujan's notebooks

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    $\begingroup$ math.stackexchange.com/questions/369846/… $\endgroup$
    – individ
    Apr 23, 2020 at 4:34
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    $\begingroup$ math.stackexchange.com/questions/1018893/when-does-x3y3-kz2/… $\endgroup$
    – individ
    Apr 23, 2020 at 4:36
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    $\begingroup$ @individ I have made my own formula some time ago for some value $n$, but letting $n\in\mathbb{Q}$ would have been much too cumbersome. The formula is:$$\big\{8(2n+1)\underbrace{(4n+1)(16n^2+20n+7)}_{(4n+2)^3-1}\big\}^3+\big\{\underbrace{(8n+5)(64n^2+56n+13)}_{(8n+4)^3-1}\big\}^3$$ $$=\big\{(32n^2+40n+11)(1024n^4+1792n^3+1248n^2+424n+61)\big\}^2$$ Thank you for revealing those much simpler formulae. $\endgroup$
    – Mr Pie
    Apr 23, 2020 at 5:22

2 Answers 2

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Ramanujan could not have had a general formula with the machinery of the time. Note that all of the listed results can be rewritten to have the same denominator $A$ in each number, so after multiplying by $A^3$ we have $$x^3+y^3=Az^2$$ where $x,y,z$ are integers.

The given solutions correspond to the following integral relations: $$23^3+1^3=2\cdot78^2$$ $$22^3-1^3=7\cdot39^2$$ $$314^3+1^3=105\cdot543^2$$ $$313^3-1^3=104\cdot543^2$$ Also note that the second term is always $\pm1$. So Ramanujan was merely looking for factorisations of incremented and decremented cubes with small squarefree parts.

From this framework other solutions arise, like $$\left(\frac47\right)^3-\left(\frac17\right)^3=\left(\frac37\right)^2$$ $$\left(\frac{31}{38}\right)^3+\left(\frac1{38}\right)^3=\left(\frac{14}{19}\right)^2$$

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  • $\begingroup$ It seems I made a typo in the fourth relation; it has been edited now and is currently true. $\endgroup$
    – Mr Pie
    Apr 23, 2020 at 5:20
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    $\begingroup$ @MrPie Done that. $\endgroup$ Apr 23, 2020 at 5:27
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We argue about equation:

$$x^3 ± y^3=A z^3$$

Let $y=t-x$ such that:

$$x^3+(t-x)^3=t^3-3t^2x+3tx^2=t[t^2-3x(t-x)]$$

Suppose $t=z^2$, then:

$$A=t^2-3xt+3x$$

This equation can have infinite solutions, for example:

$(t, A)=(25, 163)$

which gives:

$x=11, y=14, z=5$ such that:

$$11^3+14^3=163\times 5^2$$

Now suppose $A=k^3$, then we have:

$$3x^2-3xt+k^3-t^2=0$$

$$x=\frac{3t ±\sqrt{12k^3-3t^2}}{6}$$

for example:

$(t, k)=(9, 21)$ which gives $x=60, y=-51, z=3, A=21$:

$60^3-51^3=21^3\times 3^2$$\big(\frac{60}{21}\big)^3-\big(\frac{51}{21}\big)^3=3^2$

Or:

$t=676=26^2$,$A=k^3=1183^3$, $x=23829$, $y=-507$ such that:

$$\big(\frac{23829}{1183}\big)^3-\big(\frac{507}{1183}\big)^3=25^2$$

If discriminant $\Delta=9t^2-12t^2+12k^3=0$, then there are also infinite solutions, such that $x=y$, for example:

$(k, t)=(64, 1024)$ which gives $x=y=512$ and we have:

$\big(\frac{512}{64}\big)^3+\big(\frac{512}{64}\big)^3=32^2$

Which can be reduced to: $2\times 8^3=32^2$

Or:

$(t, k)=(11664=108^2, 324)$ which gives $x=y=5832$ and we have:

$\big(\frac{5832}{324}\big)^3+\big(\frac{5832}{324}\big)^3=108^2$

Which can be reduced to:

$2\times 18^3=108^2$

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  • $\begingroup$ In your first equation, I think you mean to let $t=z^3$ not $z^2$, but I still follow. Nice answer! :) $\endgroup$
    – Mr Pie
    Apr 24, 2020 at 0:39
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    $\begingroup$ In equation we have $z^2$, so I meant $t=z^2$ to be able to argue. $\endgroup$
    – sirous
    Apr 24, 2020 at 5:28
  • $\begingroup$ Oh whoops, sorry, idk why I thought that. Fermat's Last Theorem should have spoken to me sooner! $\endgroup$
    – Mr Pie
    Apr 24, 2020 at 5:28

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