Proportionality of a system of polynomials

Question

I am currently reading Ivanovs „Easy as Pi?“ in fact, I am trying to understand a proof in this book. Following statement is to prove:

For $n\geqslant 3$ the curve $x^n+y^n=1$ has no rational parametrization.

The proof goes by contradiction. For this we look at three relatively prime real polynomials $p(t),q(t),r(t)$, not all constant; so that

$(\frac{p(t)}{r(t)})^n+ (\frac{q(t)}{r(t)})^n=1$

this is equivalent to:

$p^n(t)+q^n(t)-r^n(t)=0$

by taking the derivative we derive:

$p‘(t)p^{n-1}(t)+ q‘(t)q^{n-1}(t) \ +$ $r‘(t)r^{n-1}(t)=0 $

If we now look at the system:

$p(t)x+q(t)y+r(t)z=0$ $p‘(t)x+q‘(t)y+r‘(t)z=0$

We see $x= p^{n-1}(t),y= q^{n-1}(t), z= r^{n-1}(t)$ is a solution.

Now comes the part, that I don‘t understand, from now I will write the functions without the argument

Since every solution of this system is proportional to the triple,(which is also a solution):

$(q‘r-qr‘,r‘p-rp‘,p’q-pq‘)$

We find relatively prime polynomials $f,g$, so that

$p^{n-1}=\frac{f}{g}(q‘r-qr‘)$ $q^{n-1}=\frac{f}{g}(r‘p-r‘p)$ $r^{n-1}=\frac{f}{g}(p‘q-pq‘)$

Why is every solution proportional to the triple?

My inital thought was that, if you think about a linear system of equations, the solutions to it create a vector subspace. But this can‘t be the reason, since we have multiplication of polynomials.

Answer 1

The Answer is quite simple, just don‘t overcomplicate it.

Since,

$(q‘r-qr‘,r‘p-rp‘,p’q-pq‘)$

solves the system, and

$(p^{n-1},q^{n-1},r^{n-1})$

does aswell, we know that

$p(q‘r-qr‘)+q(r‘p-rp‘)y+r(p’q-pq‘)z=0$ $p‘(q‘r-qr‘)+q‘(r‘p-rp‘)y+r‘(p’q-pq‘)z=0$

and

$p(p^{n-1})+q(q^{n-1})y+r(r^{n-1})z=0$

$p‘(p^{n-1})+q‘(q^{n-1})y+r‘(r^{n-1})z=0$

Since for some arbitrary $s\in\mathbb{R}[x]$ the triple

$(s(q‘r-qr‘),s(r‘p-rp‘),s(p’q-pq‘))=(s_1,s_2,s_3) $

also solves the system (multiply $s\neq0$ on both sides). We know for a fact that there are relatively prime $f,g\in\mathbb{R}[x]$, so that:

$(\frac{f}{g}(q‘r-qr‘),\frac{f}{g}(r‘p-rp‘), \frac{f}{g}(p’q-pq‘)) = (p^{n-1},q^{n-1},r^{n-1})$,

since both of them are solutions, this also works for every other solution.