Solution to the first-year puzzle

In this puzzle, we asked: suppose a+b=1 and a^3+b^3=1. What then is a^{17}+b^{17}?

First let us solve this by hand. Since a+b=1, (a+b)^3=1. So

(a+b)^3=a^3+3a^2b+3ab^2+b^3=1=a^3+b^3.

Hence it follows that

3a^2 b + 3ab^2=3ab(a+b)=0.

But since a+b=1, this means that ab=0, so if a=0 we must have b=1 or vice versa. In any case, it follows that a^{17}+b^{17}=1.

The reason I wanted to present this problem is that it shows the power and beauty of Gröbner bases. Here is how to solve this problem in MAPLE. Define z= a^{17}+b^{17}. Then the question is: what equation does z satisfy? So we set up the three equations, two of which are the data of the problem and the last one is the definition of z. Crank up MAPLE…

with(Groebner):
e1 := a+b-1:
e2 := a^3+b^3-1:
e3 := a^17+b^17-z:
Basis([e1,e2,e3],plex(z,a,b));
[-b + b , a + b – 1, z – 1]

The Basis([e1,e2,e3],plex(z,a,b)) command computes the Gröbner basis of (the ideal defined by) this set of equations (with the right lexicographic ordering) and has to give a univariate polynomial in z. The resulting three equations are equivalent (in terms of their zeroes) to the original ones. The only zero of z-1 is z=1.

I leave many things unexplained on purpose: if you are interested, the help system of MAPLE and all the resources of the web are at your disposal.

(MG)

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