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Thue proved that all Diophantine equations consisting of an irreducible binary form (cubic or higher) equal to a constant, i.e., $$c_nx^n+c_{n-1}x^{n-1}y+\cdots+c_oy^n=k$$ ($n,k$ fixed) have finitely many solutions. What is known about equations that are ternary rather than binary forms? I.e., $$c_{n,0}x^n+c_{n-1,1}x^{n-1}y+c_{n-1,0}x^{n-1}z+c_{n-2,1}x^{n-2}yz+\cdots+c_{0,0}z^n=k$$ with $n,k$ fixed.

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Here are a few papers about ternary forms of degree at least 3:

MR0006740 (4,34c) Mahler, Kurt Remarks on ternary Diophantine equations. Amer. Math. Monthly 49, (1942). 372–378.

MR0025489 (10,14c) Huff, Gerald B. Diophantine problems in geometry and elliptic ternary forms. Duke Math. J. 15, (1948). 443–453. [Note: this paper studies $ax(y^2-z^2)=by(x^2-z^2)$]

MR0062767 (16,14g) Selmer, Ernst S. A conjecture concerning rational points on cubic curves. Math. Scand. 2, (1954). 49–54.

MR0167460 (29 #4733) Oppenheim, A. The rational integral solution of the equation $a(x^3+y^3)=b(u^3+v^3)$ and allied Diophantine equations. Acta Arith. 9 1964 221–226.

MR0376520 (51 #12695) Dofs, Erik On some classes of homogeneous ternary cubic Diophantine equations. Ark. Mat. 13 (1975), 29–72. [Note: this paper studies $ax^3+by^3+cz^3=dxyz$]

MR1704332 (2000h:11033) Bayadilov, E. E. On the divisor problem for values of a ternary cubic form. (Russian) Vestnik Moskov. Univ. Ser. I Mat. Mekh. 1999, no. 1, 58--60; translation in Moscow Univ. Math. Bull. 54 (1999), no. 1, 43–45

MR2095257 (2005i:11037) Choudhry, Ajai Symmetric Diophantine equations. (English summary) Rocky Mountain J. Math. 34 (2004), no. 4, 1281–1298.

There is also work on $\it pairs$ of ternary forms, where degree $2$ is already interesting:

MR1378574 (96m:11026) Gaál, István(H-LAJO-IM); Pethő, Attila; Pohst, Michael(D-TUB-3) Simultaneous representation of integers by a pair of ternary quadratic forms—with an application to index form equations in quartic number fields. (English summary) J. Number Theory 57 (1996), no. 1, 90–104.

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For example, $x^3 + y^3 - 2 z^3 = 0$ has infinitely many solutions.

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Or $x^n + y^n - 2z^n = 0$, for arbitrary $n$ –  Thomas Andrews Jun 14 '11 at 19:04
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This is an old post. (Ok, a very old post.) But for the benefit of those who might come across this, the equation,

$$x^3+ny^3+n^2z^3-3nxyz = 1$$

is sometimes known as the cubic Pell equation. If for a given constant $n$, there is an integer solution $x,y,z$, then one can generate an infinite more from that initial triple.

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