## A problem with polynomials: solution (part 1)

The problem before us was: show that the equations $x^3-x-1=0$ and $x^5-x^4-1=0$ have a root in common. I know that we asked you to show that they have a real root in common, but the method of solution will indeed demonstrate that every root of the cubic is also the root of the quintic. I will do this by hand, and then I will show you the MAPLE proof.

So let $x$ be any root of the cubic. Then $x^3=x+1$. But then $x^4=x^2+x$ and $x^5=x^3+x^2$. So

$x^5-x^4-1=(x^3+x^2)-(x^2+x)-1=x^3-x-1=0,$

since $x$ was a root of the cubic. So $x$ is also a root of the quintic. C’est tout. In MAPLE you would do the following:

f1 := x^3-x-1:
f2 := x^5-x^4-1:
resultant(f1,f2,x);

0

This is good, as this shows (why?) that the two polynomials have a root in common. Let us now find how many roots they have in common…

with(linalg):
A := sylvester(f1,f2,x):
nops(kernel(A));

3

So the answer is three roots in common (why?)! Don’t tell me this is not neat. For now I will leave it to you to work out what exactly I did there — the code is strewn with clues. Please also explain to yourself why this is a proof.

(MG)