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)