This is one of these wonderful results you can get with only pencil and paper, a lot of time (better with a good computer algebra system), and a very good idea. The author considers the general quartic form in two variables and rational coefficients \[ f(x,y)= ax^4+ bx^3y+ cx^2y^2+ dxy^3+ ey^4 \] and asks for rational solutions of \(f(x,y)= f(u,v)\). Theorem: Take \(q(m,n)= 4amn^3- 3bmn^2+ 2cmn- dm- bn^3+ 2cn^2- 3dn+ 3e\). A necessary and sufficient condition that the quartic Diophantine equation \(f(x,y)= f(u,v)\) has an integer solution with \(y\neq\pm v\) is that there exist rational numbers \(n, m\) such that \(q(m,n)\) and \(q(n,m)\) are both zero or \(-q(m,n) q(n,m)\) is a non-zero perfect square. The proof is by explicitly writing down a solution, if the conditions are satisfied, or by construction of \(n\) and \(m\), if a solution is known. The author gives two examples not covered by the so far only relevant theorem on this question by \textit{B. Segre} [Proc. Lond. Math. Soc. (2) 49, 353--395 (1947; Zbl 0034.08603)]. In both cases the respective \(m\) and \(n\) are found by trial. Reviewer's remarks: There seems to be no theory behind finding these \(m, n\). E.g., L. Euler has given the solution \(x=158\), \(y=59\), \(u=134\), \(v=133\) of \(x^4+ y^4= u^4+ v^4\) [cf. \textit{L. E. Dickson}, ``History of the theory of numbers. Vol. II: Diophantine analysis'', reprint New York (1971); Chap. XXII: Equations of degree four, p. 646] by a general method [cf. also \textit{L. J. Mordell}, Diophantine equations, Pure Appl. Math. 30, London etc.: Academic Press (1969; Zbl 0188.34503); Chap. 12: Rational and integer points on quartic surfaces, p. 91]. You can compute afterwards that \(m=12/37\), \(n=-73/48\), but how to find this a priori? So we have a nice result but you will be lucky to get a practical solution. Thinking of Segre, perhaps it would be helful to give the geometric background of the author's method.