Guess Diophantine equation from its solutions

Question

Take for example the equation $$ n^2=x^2+y^2+1 $$ where $x,y$ are integers.

This equation admits solutions only for some values of $n$, to wit, $$ n=1,3,9,17,19,33,35,51,73,81,99,\dots $$

Can we reverse-engineer the Diophantine equation from the sequence above? I guess the solution is non-unique, but any solution will do.

Answer 1

Although perhaps not very satisfying, a Diophantine equation yielding the values of n in the question is given by

lst = {1, 3, 9, 17, 19, 33, 35, 51, 73, 81, 99};
eq = n == Sum[lst[[i]] x[i], {i, 11}]
(* n == x[1] + 3 x[2] + 9 x[3] + 17 x[4] + 19 x[5] + 33 x[6] + 35 x[7] + 
        51 x[8] + 73 x[9] + 81 x[10] + 99 x[11] *)

The first eleven non-trivial solutions are

tab = {n, Table[x[i], {i, 11}]} // Flatten;
Table[tab /. {n -> lst[[i]], x[i] -> 1, x[_] -> 0}, {i, 11}]
(* {{1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}, {3, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0}, 
    {9, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0}, {17, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0}, 
    {19, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0}, {33, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0}, 
    {35, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0}, {51, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0}, 
    {73, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0}, {81, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0}, 
    {99, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1}} *)

In fact, there is an infinite number of Diophantine equations satisfying the requirements of the question. Obtaining a particular equation would require several constraints.