For any even $n$, there should be a polynomial $f(x,y)$ that vanishes on the points $(\cos^n(t),\sin^n(t))$ for all $t$ (since it is the image of the projective variety $x^2+y^2 = 1$ under the $n/2$thpowerineachcoordinate map). Is it possible to give an explicit form for $f$?

$\begingroup$ Your question is unclear: $\cos^n(t)$ is not a point of the real plane, and nor is $\sin^n(t)$. Do you mean that you want a polynomial $f$ such that $f(\cos^n(t),\sin^n(t))=0$ for all $t$ ? and i think you mean "under the $n/2$^th power... $\endgroup$ – GreginGre Sep 20 '19 at 13:49

$\begingroup$ Yes, sorry, fixed these things! $\endgroup$ – user142054 Sep 20 '19 at 13:51

$\begingroup$ See superellipse. $\endgroup$ – Lucian Sep 26 '19 at 10:02
I don't know if this is irreducible, but it gives an answer: Let $\zeta$ be a primitive $n/2$root of unity. Multiply out $$\prod_{a=1}^{n/2} \prod_{b=1}^{n/2} (\zeta^a x^{2/n} + \zeta^b y^{2/n}  1 ).$$ The result is a polynomial in $x$ and $y$, and one of the factors is $x^{2/n} + y^{2/n}1$, so it vanishes when $(x,y) = (\cos^n t, \sin^n t)$.

2$\begingroup$ Thanks! It looks like it has the correct degree for the image curve $(n/2)$, so it ought to be irreducible. How do I see that it's a polynomial? $\endgroup$ – user142054 Sep 20 '19 at 14:34

4$\begingroup$ The Galois group of $\mathbb{C}(x^{2/n}, y^{2/n})$ over $\mathbb{C}(x,y)$ is $\mathbb{Z}/(n/2)^2$, given by $(x,y) \mapsto (\zeta^a x, \zeta^b y)$. The formula David wrote down is just $x^{2/n} + y^{2/n}  1$ multiplied by all its Galois conjugates, so this expression is manifestly invariant under the Galois group. So in fact it's an element of $\mathbb{C}(x,y)$. (Degree considerations then tell you it's a polynomial and not just a rational function) $\endgroup$ – Kevin Casto Sep 20 '19 at 20:06

$\begingroup$ @KevinCasto Thanks! You can also see that it is a polynomial because $\mathbb{C}[x,y]$ is integrally closed, and $x^{2/n}$ and $y^{2/n}$ are manifestly integral over it. $\endgroup$ – David E Speyer Sep 23 '19 at 13:19

2$\begingroup$ Its pullback along the $2/n$th power map has $(n/2)^2$ irreducible components, given by the linear factors of your equation, which are all distinct since the linear factors have the same constant terms but different linear terms, and the Galois group acts transitively on them, so indeed your polynomial must be irreducible. $\endgroup$ – Will Sawin Sep 24 '19 at 19:51
You can use polynomial elimination, which is implemented in Macaulay2.
In fact, the parametric equations of your affine curve are $$x=\frac{(2t)^n}{(1+t^2)^n}, \quad y=\frac{(1t^2)^n}{(1+t^2)^n},$$ so that an implicit equation $f(x, \,y)=0$ for it is obtained by eliminating the variable $t$ among these, i.e. eliminating $t$ in the ideal $$I=(x(1+t^2)^n(2t)^n, \, y(1+t^2)^n(1t^2)^n) \subset \mathbb{Q}[x, \, y, \, t].$$ We can do this for all the values of $n$, non only the even ones. For instance, in the case $n=5$ the script is the following:
R=QQ[x, y, t];
n=5;
I=ideal(x*(1+t^2)^n(2*t)^n, y*(1+t^2)^n(1t^2)^n);
J=eliminate(t, I);
Now the command "gens J" gives a list of generators for the elimination ideal, that consists only of the polynomial we are looking for:
gens J
 x10+5x8y2+10x6y4+10x4y6+5x2y8+y105x8+605x6y21905x4y4+605x2y65y8+10x6+1905x4y2+1905x2y4+

10y610x4+605x2y210y4+5x2+5y21 
For even $n$, the generator of $J$ has degree $n/2$, as expected. For example, when $n=6$ the script produces the following polynomial of degree $3$:
gens J
 x3+3x2y+3xy2+y33x2+21xy3y2+3x+3y1 

$\begingroup$ This is nice, thanks! Either way it looks like it is going to be hard to compute a specific coefficient. $\endgroup$ – user142054 Sep 20 '19 at 14:37

$\begingroup$ The answer should be degree $5.$ In fact this is correct if you change $x^2$ and $y^2$ to $x$ and $y$, $\endgroup$ – Aaron Meyerowitz Sep 21 '19 at 6:21

$\begingroup$ @AaronMeyerowitz: why the elimination procedure does not give me the right degree? $\endgroup$ – Francesco Polizzi Sep 21 '19 at 6:34

$\begingroup$ I'm not sure. Perhaps it is actually $x^2=\frac{(2t)^n}{(1+t^2)^n} ?$ But the degree should be $5$ and I did check that it works. See my answer below for alternate ways to write the polynomial. $\endgroup$ – Aaron Meyerowitz Sep 21 '19 at 6:51

1$\begingroup$ You have $x$ and $y$ switched, not that it matters. OK! I see now! The stated problem is for $n$ even. So actually you are right. Your degree $10$ polynomial satisfied by all the points $(x,y)=(cos^5(t),sin^5(t))$ is even. If we halve the degrees it is degree $5$ and satisfied by all the points $(x,y)=(cos^10(t),sin^10(t)).$ $\endgroup$ – Aaron Meyerowitz Sep 21 '19 at 8:21
Here are some small explicit solutions and general comments.
Since the polynomials are symmetric, one would expect them to be more nicely expressed using $$S=x+y \\ M=xy\\ P=xy.$$
Here $M$ should only appear to even powers and $M^2=(xy)(yx).$
It seems convenient to use all three although either $P$ or $M^2$ suffices with $S$ since
$4P=S^2M^2.$
Also, I will bend notations and introduce a symbol $T$ with $T^k$ interpreted as $M^k$ or $S^k$ according as $k$ is even or odd.
 For $n=2$ the desired polynomial is , of course, $$S1=0.$$
 For $n=4$ the desired polynomial is $$(S1)^24P= \\ (T1)^2=0$$
 For $n=6$ the desired polynomial is $$(S1)^3+27P= \\ (T1)^3+15P=0$$
 For $n=8$ the desired polynomial is $$(S1)^48P\left(S^22P+14S+17\right)= \\(T1)^4112P(S+1)=0$$
 For $n=10$ I was unable to derive the polynomial (see comments below) however the degree $10$ polynomial given by Francesco satisfied by the points $(x,y)=(\cos^5(t),\sin^5(t))$ is even. If we halve the degrees it is degree $5$ and satisfied by all the points $(x,y)=(\cos^{10}(t),\sin^{10}(t)).$
${x}^{5}+5\,{x}^{4}y+10\,{x}^{3}{y}^{2}+10\,{x}^{2}{y}^{3}+5\,x{y}^{4}+ {y}^{5}5\,{x}^{4}+605\,{x}^{3}y1905\,{x}^{2}{y}^{2}+605\,x{y}^{3}5 \,{y}^{4}$
$+10\,{x}^{3}+1905\,{x}^{2}y+1905\,x{y}^{2}+10\,{y}^{3}10\,{x }^{2}+605\,xy10\,{y}^{2}+5\,x+5\,y1 $
$$(S1)^5+625P(M^2+3S+1P)=\\(T1)^5+15P(39(T1)^2+203S47P))=0$$
The cases of $n=10,12,16$ can be worked out and continue to increase in complexity at about the same rate.
The curve we want is $x^{1/m}+y^{1/m}=1$ where $n=2m.$ So we could as well think of this as an image of the` curve $x+y=1.$ Really we only want the restriction to the positive quadrant. For $m$ even the equation is only defined in the positive quadrant. This is not the case for $m$ odd. Perhaps the right thing to do is to think of it as the unit ball in the $L_p$ metric for $p=\frac1m$: $$u^{1/m}+v^{1/m}=1.$$
For $n=4,6,8$ it isn't to bad to start from $x^{1/m}+y^{1/m}=1$ and manipulate be rearranging and raising to powers to get a polynomial. I'm not sure that continues.
Instead I used David Speyer's method. The primitive $m$th roots of unity have reasonably simple expressions for $m=2,3,4,6,8.$ Also $m=5$ is nice but I couldn't quite get that to work.
What about a polynomial satisfied by the points $(\sin^n(t),\cos^n(t))$ for $n$ odd? Just take the polynomial for $(\sin^{2n}(t),\cos^{2n}(t))$ and double all the exponents.