# Varieties with few monomials and the n-conjecture

The n-conjecture is a generalization of abc and basically says that the if $a_1 + \ldots + a_n=0$, no proper subsum vanishes and $a_i$ are coprime, then the radical of $a_1\cdots a_n$ can't be too small.

Let $f(x_1,\ldots,x_m)=0$ be possibly reducible projective variety.

Take $a_i$ in the n-conjecture to be the monomials in $f$ (including coefficients).

If $f$ has $n$ monomials, the monomials are coprime at infinitely many integral points, no subsum vanishes and there isn't great discrepancy in the size of the monomials the n-conjecture implies the degree of $f$ is at most $(2n-5)m$.

This bound is sharp:

$$x^9+y^9+3ax^3y^3z^3-a^3z^9=\left(-1\right) \cdot (z^{3} a - x^{3} - y^{3}) \cdot (z^{6} a^{2} + x^{3} z^{3} a + y^{3} z^{3} a + x^{6} - x^{3} y^{3} + y^{6})$$

Both factors of the RHS are genus $1$.

Let $f(x,y,z)=b_1 x^d+b_2 y^d+b_3 z^d+b_4 x^{d_1}y^{d_2}z^{d_3}$ and all $b_i$ are nonzero.

Q1 Can $f(x,y,z)=0$ have genus $0$ or $1$ component if $d>9$ ?

Q2 Does infinitely many coprime integral points on varieties with few monomials contradict other conjectures?

For positive even $k$ up to $40$ all of these are genus $1$ and similar are of positive rank.
$$x^k-y^2(x^4+1)=0$$
$$x^k-y^2(x^4+x+1)=0$$
$$(x^4-1)^k-y^2(x^4+x+1)=0$$