Re: 61. - THE SILVER CUBES

In article <104uq97$1b42r$1@artemis.inf.ed.ac.uk>,
Richard Tobin <richard@cogsci.ed.ac.uk> wrote:

I had to look up how you might go about it.  It turns out
>
(1) There's a solution with small numerators and denominators, but
   one of them is negative;
>
(2) There's a recurrence relation that given one solution produces
   another.
>
Applying (2) to (1) gives a solution in positive rationals.
>
So an easier puzzle is to find (1).

Spoiler
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The simple but invalid solution is (18/7)^3 + (-1/7)^3 = 17

Given (x/z)^3 +(y/z)^3 = a, the recurrence relation gives a new triple
x,y,z:

x(x^3 + 2y^3)
-y(2x^3 + y^3)
z(x^3 - y^3)

Applying it to 18,-1,7 gives the desired solution:

(11663/40831)^3 + (104940/40831)^3 = 17

which can also be find in a reasonable time by a computer program.

Applying the relation again gives another solution with a negative
cube, but applying it a third time gives another valid solution:

(213911660725112665205915686283667768508559220489375013086586592912683129002901680/85663575208595989523064839018954307150548744631241583862215046886695340878091183)^3 +
(96489450149187273559455691876125510797936653514955100397233219218908577035631359/85663575208595989523064839018954307150548744631241583862215046886695340878091183)^3 = 17

-- Richard