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I need to find $x$ such that $P(x)=x^6-2x^5+2x+1=0$ and $Q(x)=3x^5-5x^4+1=0$

Of course we know that the common roots of two polynomials are the roots of their gcd, so we could simply compute the gcd.

But i found another method: we make "combinations" of the equations in order to reduce the powers of x (method of successive subtraction):

So, if $P(x)=0$ and $Q(x)=0$ then we have $xQ(x)-3P(x)=0$ so $P'(x)=x^5-5x-3=0$

Then we have $3P'(x)-Q(x)=0$ so $Q'(x)=x^4-3x-2=0$

Finally it would follow that $P'(x)-xQ'(x)=0$ so $x^2-x-1=0$

The computations are simple but I'm interested in the logic, more exactly in the "equivalence"/reversibility of these computations.

Logically, clearly if x verifies $P(x)=Q(x)=0$ then $x$ must verify the last equation $x^2-x-1=0$. So if there exists a solution , it is a solution to this last equation. But how do i know for sure that the solutions of this last equation are solution to the first ones??? I mean, could i use this method and be wrong, so that not all the solutions of the last equation are solutions of the first? How do i know this "algorithm" stops at the right equation? For example if i would have "stopped" at $Q'(x)$ and could have solved it completely (even if it is 4-th degree) not all the solutions would be solutions for both the initial polynomials. What am i missing here?

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  • $\begingroup$ Note that what you computed is the gcd of P and Q: $gcd(P, Q) = x^2-x-1$. What you did is similar to the euclidean algorithm, only with differences instead of remainders. $\endgroup$ – Martin R Mar 8 at 7:54
  • $\begingroup$ But how do i show this is the gcd? In euclidean alg the "termination condition" is when we obtain a zero remainder. Here with differences what is the stopping condition? $\endgroup$ – amarius8312 Mar 8 at 13:18

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