Are there infinitely many triples of integers $(x,y,z)$ satisfying the equation $$ x^4+y^3+z^2+1=0 \quad ? $$ This is one of the simplest-looking equations (another one is famous $x^3+y^3+z^3=3$) for which this question is open. It is left open in my book https://doi.org/10.1007/978-3-031-62949-5 , see also arxiv preprint https://arxiv.org/abs/2404.08518 .
Similar equations of the form $x^4+y^3=z^2+n$ has been discussed in the previous question Representing integers as sums of three powers , but, if I remember correctly, this specific equation has not been explicitely asked on Mathoverflow yet.
Examples of solutions are
x = 0, y = -1, z = 0
x = 1, y = -3, z = 5
x = 6, y = -13, z = 30
x = 11, y = -27, z = 71
x = 31, y = -147, z = 1501
x = 51, y = -203, z = 1265
x = 29, y = -5507, z = 408669
x = 1386, y = -15661, z = 388458
x = 288, y = -16957, z = 2206566
x = 1669, y = -21531, z = 1490663
x = 2239, y = -29931, z = 1297207
x = 855, y = -38051, z = 7386395
x = 2761, y = -47171, z = 6844587
x = 2237, y = -57147, z = 12711719
x = 1103, y = -84027, z = 24326849
x = 2859, y = -85547, z = 23648369
x = 949, y = -90891, z = 27387137
and so on.
Because the exponents are $(4,3,2)$, and $1/4+1/3+1/2>1$, heuristic strongly suggest infinity of solutions, but how to prove this?
