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arxiv: math/0505629 · v2 · submitted 2005-05-29 · 🧮 math.HO · math.NT

Observations on two fourth powers whose sum is equal to the sum of two other fourth powers

classification 🧮 math.HO math.NT
keywords eulergivessolutionothersectbiquadrataconjecturefourth
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Translated from the Latin original, "Observationes circa bina biquadrata quorum summam in duo alia biquadrata resolvere liceat" (1772). E428 in the Enestroem 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 nontrivial integer solutions to $A^4+B^4+C^4=D^4$. I do not know whether he stated this conjecture previously. In sect. 3, Euler sets A=p+q, B=p-q, C=r+s,D=r-s. Taking r=p and s=q gives the trivial solution of C=A and B=D, but this gives Euler the idea of making p and r multiples of each other and q and s multiples of each other. If k=ab this again 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. 377-380 gives a faithful explanation of Euler's solution.

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