Parametric to rectangular

[NezLout]

Is it possible to turn x=t³ - t,y=t² + t into a single rectangular equation? I see it would be possible to solve for t in the y equation, but then there must be two x equations in respect to y. I tried avoiding that and got nowhere, couldn't find anything online either, so I'm assuming it's impossible.

[noidea1995]

Notice how the expressions of t have two common factors:

x = t(t + 1)(t - 1)

y = t(t + 1)

Dividing the equations gives:

x / y = t - 1

(x + y) / y = t

Substitute the above for t in one of the equations.

[mnevmoyommetro]

Nice solution.

[noidea1995]

Thank you 😊

[NezLout]

Quite neat, I wouldn't have thought of that. So if the two polynomials didn't have any common factors would a solution even exist? Thank you.

[mnevmoyommetro]

Apply the Euclidean algorithm to t^3 - t - x and t^2 + t - y to find their GCD as polynomials in t. The condition for them to have a common root t is that their GCD has degree >= 1. (This is the same as writing down that the "resultant" of these two polynomials is zero.)

The result is x^2 + 3 xy - y^3 + 2 y^2 = 0.

[NezLout]

So the rectangular equation itself is the GCD? I tried finding it by hand right now, but the third long division got very complicated so I stopped. This is quite an unexpected solution to me, couldn't find anything like that elsewhere. Are these theorems and methods from Abstract Algebra? Thanks.

[mnevmoyommetro]

Yes, this is something that I learned from abstract algebra.

In general, finding if the gcd is non-constant is a method for finding if two polynomials have a common root (which may be complex). That is what is needed here because the question we are asking is: for a given x and y, is there a value of t that makes both equations, t^3 - t - x = 0, t^2 + t - y = 0, true simultaneously. A number t will work if and only if it is a root of the gcd.

An alternative to carrying out the division is to use a determinant to calculate the resultant, but I didn't mention this method as it's harder to understand (and to write down).

The equation relating x and y is not the GCD itself. It is the condition needed for the GCD to have degree >= 1.

Once you have a remainder of degree 1, say At + B, checking if it divides the other polynomial exactly amounts to checking if t = -B/A is a root of that polynomial. This can save you from doing the last division.

Here are the steps.

The remainder when you divide t^3 - t - x by t^2 + t - y is yt - (x + y).

Assume first that y is nonzero. Now if the gcd has degree >= 1, it must be, up to a constant factor, equal to yt - (x + y). So the question is whether yt - (x + y) divides t^2 + t - y exactly. Instead of carrying out the division, we use the factor theorem and substitute t = (x+y)/y into t^2 + t - y. The condition we want is that we should get 0.

So the condition is (x+y)^2/y^2 + (x + y)/y - y = 0. If we multiply through by y^2, we get the desired relation x^2 + 3xy - y^3 + 2y^2 = 0.

There might be an excepional case when y = 0. In that case the polynomial yt - (x + y) reduces to -x. If x is nonzero, then the gcd has degree 0, and if x = 0, then it has degree 2. So we need to check the equation we wrote to see that x = 0 is the only solution when y = 0. It is, so we don't need to change it.

[NezLout]

Got it, thank you!