# 87 arithmetic/sum.of.cubes.p

## Description

This article is from the Puzzles FAQ,
by Chris Cole chris@questrel.questrel.com and Matthew Daly
mwdaly@pobox.com with numerous contributions by others.

# 87 arithmetic/sum.of.cubes.p

Find two fractions whose cubes total 6.

arithmetic/sum.of.cubes.s

Restated:

Find X, Y, minimum Z (all positive integers) where

(X/Z)^3 + (Y/Z)^3 = 6

Again, a generalized solution would be nice.

You are asking for the smallest z s.t. x^3 + y^3 = 6*z^3 and x,y,z in Z+.

In general, questions like these are extremely difficult; if you're

interested take a look at books covering Diophantine equations

(especially Baker's work on effective methods of computing solutions).

Dudeney mentions this problem in connection with #20 in _The Canterbury

Puzzles_; the smallest answer is (17/21)^3 + (37/21)^3 = 6.

For the interest of the readers of this group I'll quote:

"Given a known case for the expression of a number as the sum or

difference of two cubes, we can, by formula, derive from it an infinite

number of other cases alternately positive and negative. Thus Fermat,

starting from the known case 1^3 + 2^3 = 9 (which we will call a

fundamental case), first obtained a negative solution in bigger

figures, and from this his positive solution in bigger figures still.

But there is an infinite number of fundamentals, and I found by trial

a negative fundamental solution in smaller figures than his derived

negative solution, from which I obtained the result shown above. That

is the simple explanation."

In the above paragraph Dudeney is explaining how he derived (*by hand*)

that (415280564497/348671682660)^3 + (676702467503/348671682660)^3 = 9.

He continues:

"We can say of any number up to 100 whether it is possible or not to

express it as the sum of two cubes, except 66. Students should read

the Introduction to Lucas's _Theorie des Nombres_, p. xxx."

"Some years ago I published a solution for the case 6 = (17/21)^3 +

(37/21)^3, of which Legendre gave at some length a 'proof' of

impossibility; but I have since found that Lucas anticipated me in

a communication to Sylvester."

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