If n is odd, it is more natural to consider all integers, rather than just positive ones. Are the following the only known solutions for n = 5 ? 133^5+110^5+84^5+27^5-144^5 = 0 14068^5+6237^5+5027^5-14132^5-220^5 = 0 85282^5+28969^5+3183^5+55^5-85359^5 = 0 due to Lander & Parkin, Scher & Seidl, Frye. In connexion with a recent comment: Sir Henry Peter Francis Swinnerton-Dyer, 16th Baronet, was born 2 August 1927 and so was 16 (in fact, 15? -- possibly 14 when it was written?) when the following was published: MR0009027 (5,89e) Dyer, P. S. A solution of $A\sp 4+B\sp 4=C\sp 4+D\sp 4$. J. London Math. Soc. 18, (1943). 2--4. Euler gave solutions of the Diophantine equation $A^4+B^4=C^4+D^4$. A simpler solution was obtained by Gérardin [Intermédiaire des Math. 24, 51 (1917)]. In both solutions $A$, $B$, $C$, $D$ are homogeneous polynomials in two variables of degree seven with integral coefficients. The author gives a very simple proof for Gérardin's result. Reviewed by A. Brauer On Mon, 19 Jun 2006, Gary McGuire wrote:
Christian Boyer wrote:
Jean-Charles Meyrignac thinks that a^n + b^n = c^n + d^n, with n>4, is impossible.
On a^5 + b^5 = c^5 + d^5 + e^5, he says that the ONLY known solution is 14132^5 + 220^5 = 14068^5 + 6237^5 + 5027^5 = 563661204304422162432 So, for sure, it will be incredibly difficult to find solutions with e=0.
The first counterexample to a conjecture of Euler was
84^5+110^5=144^5+(-27)^5+(-133)^5.
Do you want solutions in the integers or in the positive integers?
Gary McGuire
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