Translated from the Latin original, "Observationes circa bina biquadrataquorum summam in duo alia biquadrata resolvere liceat" (1772). E428 in theEnestroem index. This paper is about finding A,B,C,D such that $A^4+B^4=C^4+D^4$. In sect. 1,Euler states his "quartic conjecture" that there do not exist any nontrivialinteger solutions to $A^4+B^4+C^4=D^4$. I do not know whether he stated thisconjecture previously. In sect. 3, Euler sets A=p+q, B=p-q, C=r+s,D=r-s. Taking r=p and s=q givesthe trivial solution of C=A and B=D, but this gives Euler the idea of making pand r multiples of each other and q and s multiples of each other. If k=ab thisagain gives the obvious solution, so in sect. 5: Perturb k to be $k=ab(1+z)$. Euler works out two solutions. One is A=2219449, B=-555617, C=1584749,D=2061283. Hardy and Wright, fifth ed., p. 201 give a simpler parametric solution of$A^4+B^4=C^4+D^4$. Thomas Heath in his Diophantus, second ed., pp. 377-380gives a faithful explanation of Euler's solution.
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机译：源自拉丁文原著，《二重奏中的双峰观察》，另译为“ biquadrata resolvere liceat”（1772年）。雌激素指数中的E428。本文是关于找到A，B，C，D使得$ A ^ 4 + B ^ 4 = C ^ 4 + D ^ 4 $。在宗派。 1，Euler陈述了他的“四次猜想”，即对$ A ^ 4 + B ^ 4 + C ^ 4 = D ^ 4 $不存在任何非平凡的解答。我不知道他以前是否说过这个猜想。在宗派。在图3中，欧拉集A ＝ p + q，B ＝ p-q，C ＝ r + s，D ＝ r-s。取r = p和s = q给出C = A和B = D的平凡解，但这给了Euler想法，使pand成为彼此的倍数，q和s成为倍数。如果k = ab，这又给出了明显的解决方案，所以在节中。 5：将k扰动为$ k = ab（1 + z）$。欧拉提出了两种解决方案。一个是A = 2219449，B = -555617，C = 1584749，D = 2061283。 《哈代和怀特》，第五版，p。 201给出$ A ^ 4 + B ^ 4 = C ^ 4 + D ^ 4 $的更简单的参数解。托马斯·希思（Thomas Heath）在《丢番图》（第二版）377-380页中对欧拉的解决方案进行了忠实的解释。
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