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Let $n\in\mathbb{N}_+$.

Can each root equation of $n$ unknowns be rearranged to a polynomial equation of $n$ unknowns whose solution set contains the solution set of the root equation?

If not, is this true at least for root equations of one or two unknowns?

I already know the solution method with raising both sides of the root equation to the same power. But I guess this method is limited to simple root equations because raising to a power of a sum of at least three summands on one side of the equation does not reduce the number of roots on this side of the equation. Take e.g. the equation $\sqrt{A(x)}+\sqrt{B(x)}+\sqrt{C(x)}=\sqrt{D(x)}+\sqrt{E(x)}$.

I already know the method of introducing new unknowns. Does this method answer my question?

IV_
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    Can you clarify what an "irrational equation" is? E.g. is $ x = \pi$ an irrational equation? – Calvin Lin Dec 24 '20 at 00:51
  • If you're asking if it's always possible to eliminate roots from an equation, it is. You could write your sample equation as a non-linear system $$a+b+c=d+e \qquad a^2=A(x) \qquad b^2 = B(x) \qquad \cdots$$ and then hit it with, say, the method of resultants or Gröbner bases. (This may-or-may-not be what you mean by "the method of introducing new unknowns".) These processes are best undertaken by a computer algebra system like Mathematica or Maple, as the equations generated can explode in size. – Blue Dec 24 '20 at 13:52
  • @IV_: "We have to distinguish the explicit algebraic numbers and the implicit algebraic numbers." ... I don't know what you mean by this. Nevertheless, it's true that applying resultants or Gröbner basis to the system I mentioned will allow you to generate from $\sqrt{A(x)}+\sqrt{B(x)}+\sqrt{C(x)}=\sqrt{D(x)}+\sqrt{E(x)}$ a polynomial in $A(x)$, $B(x)$, $C(x)$, $D(x)$, $E(x)$, with the latter having all of the roots of the former (and likely many extraneous roots). These methods do for non-linear systems what Gaussian elimination does for linears. I'll write-up your example in an answer. – Blue Dec 24 '20 at 14:37
  • @Blue It's not necessary to give calculations for my example. I'm asking for the general case. If you are sure, it's enough if your answer answers my question positively. – IV_ Dec 24 '20 at 15:49
  • @IV_: Then it might be easier for me just to suggest that you do a site search for user:409 resultant. You'll see many examples of how I have invoked the method of resultants to handle various non-linear systems. The first hit is an answer where I walk-though the process; most times I just mention the method in passing. (Of course, lots of other users use the method, and Gröbner bases, so a more-general site search can be helpful, too.) Let me know if that's what you're after. – Blue Dec 24 '20 at 16:04
  • @Blue We should present here a yes/no-answer for the general case - for future readers. – IV_ Dec 24 '20 at 16:09

1 Answers1

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By irrational equation, I assume you meant one with one or more variable under a radical as assigned here. The technique of raising both sides to the same power may be invoked repeatedly if needed until there are no radicals. Take, for example.

$$A=\sqrt{x}+\sqrt{y}\implies A^2=\big(\sqrt{x}+\sqrt{y}\big)^2 = 2 \sqrt(x) \sqrt(y) + x + y$$

$$A^2 = 2 \sqrt(x) \sqrt(y) + x + y\implies A^2-x-y=2\sqrt{x}\sqrt{y}$$

$$\big(A^2-x-y\big)^2=\big(2\sqrt{x}\sqrt{y}\big)^2\implies A^4 - 2 A^2 x - 2 A^2 y + x^2 + 2 x y + y^2 = 4 x y$$

$$A^4 - 2 A^2 x - 2 A^2 y + x^2 - 2 x y + y^2 = A^4-2A^2(x+y)+(x-y)^2=0$$ The last is an algebraic equation with no radicals an no loss of the "solutions" of the original. The example has been limited to two unknowns and of the same $[(1/2)]$ power. We can, to an extent, also mix the powers of the radicals as in:

$$A=\sqrt{x}+\sqrt[3]{y}\implies (\sqrt[3]{y})^3=\big(A-\sqrt{x}\big)^3 =A^3 - 3 A^2 (x)^{1/2} + 3 A x - x^{3/2}$$ $$\implies y = A^3-3Ax-\sqrt{x}(3A^2+x)$$

$$\big(\sqrt{x}(3A^2+x)\big)^2=\big(A^3-3Ax-y\big)^2$$

$$\implies 9 A^4 x + 6 A^2 x^2 + x^3=A^6 - 6 A^4 x - 2 A^3 y + 9 A^2 x^2 + 6 A x y + y^2$$ $$\implies A^6 - 15 A^4 x - 2 A^3 y + 3 A^2 x^2 + 6 A x y - x^3 + y^2 = 0$$

poetasis
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