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It is known that the diophantine equation $x^5+y^5=z^5+t^5$, has infinity many trivial solutions. Has this equation been proven to have other integer solutions?

user26857
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Alpha
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    Such numbers are called taxibac numbers , for exponent $5$ or more , no such taxicab number is known. – Peter Dec 05 '23 at 23:05
  • euh , NO , no solutions are known apart from $x=z,y=t$. – mick Dec 05 '23 at 23:09
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    https://mathworld.wolfram.com/DiophantineEquation5thPowers.html – mick Dec 05 '23 at 23:13
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    @mick The case $p=5$ is solvable by Gaussian integers, in reference to your question on $p=11$ and Eisenstein integers. – Tito Piezas III Dec 06 '23 at 14:47
  • @TitoPiezasIII I am somewhat aware of that. I do not know the parametrization of all solutions if there is one. But I know there are solutions. Not so surprising since solutions for $p=5$ might even have ordinary integer solutions( that is an open problem) and many arguments exist for it, and those arguments get stronger for a bigger ring ofc. – mick Dec 06 '23 at 19:25
  • @TitoPiezasIII It would be nice if you had a link to all those solutions though :) Ofcourse you have the answer you give, what is nice, but I assume more are known ? – mick Dec 06 '23 at 19:26
  • @mick Yes, for $p=5$, more is known and solving it in Gaussian integers is just a special case. The degree is low enough that it can be solved by methods similar to $p=3$, and a variety of quadratic integers, but it seems just out of reach of the ordinary integers. Yes, it is an open problem, but I'm more inclined to believe the LPS conjecture. – Tito Piezas III Dec 07 '23 at 04:10
  • @TitoPiezasIII LPS is for ordinary integers only as I recall ? – mick Dec 07 '23 at 12:06

1 Answers1

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I. The OP asks for "...other integer solutions". We will interpret this to allow for Gaussian integers $a+bi$, an important class of quadratic integers. Hence, a non-trivial solution is,

$$(a+ci)^5+(b-ci)^5 = (a-ci)^5+(b+ci)^5$$

where $a^2+b^2 = 2c^2,\,$ and $a\neq b.\,$ A version was first found by Desboves. For example, $(a,b,c) = (1,7,5)$, so,

$$(1+5i)^5+(7-5i)^5 = (1-5i)^5+(7+5i)^5$$

and infinitely more.

II. However, if the OP wishes for plain vanilla integers, then no non-trivial solutions are known to,

$$a^5+b^5 = c^5+d^5 = N$$

where $N < 1.02\times10^{26}$ (Guy, 1994) cited in Mathworld. However, this has been updated to $N < 4.01\times10^{30}$ (Ekl, 1998). This roughly implies there are no solutions with both $(a,b)<10^6.$

P.S. Ekl's 25-year-old bound seems to need an update, though the Lander-Parkin-Selfridge conjecture assumes there are none anyway.

See https://en.wikipedia.org/wiki/Lander,_Parkin,_and_Selfridge_conjecture

mick
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Tito Piezas III
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