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A Pythagorean triple consists of three positive integers $a,b,c$ such that $$a^2+b^2=c^2.$$ There are infinitely many solutions. On the other hand, it is known that $$a^4+b^4=c^4$$ has no positive integer solutions.

However, I tried to modify the last equation and came up with the following question:

Are there positive integers $a,b,c,d$ such that they are all distinct and $$a^4+b^4=c^4+d^4?$$ If so, does it have infinitely many solutions? Does this equation even have a name? If so, a reference would be useful.

Bill Dubuque
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japjap
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  • Yes, just put $a=c$ and $b=d$. – Sergei Nikolaev Aug 31 '23 at 20:16
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    @SergeiNikolaev Check the edited post. – japjap Aug 31 '23 at 20:18
  • Did you try to write a program (Python, c, etc.) to check if any $a,b,c,d$ from $[1, 1000]$ fit the equation? – Sergei Nikolaev Aug 31 '23 at 20:20
  • Do you have any thoughts or working to start us off with? – H. sapiens rex Aug 31 '23 at 20:20
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    Leo. Euler said $59^4 + 158^4 = 133^4 + 134^4$. – Hanno Aug 31 '23 at 20:24
  • Does this answer your question? Expressing an Integer as the Sum of Two Fourth Powers in More than 1 way - found using an Approach0 search. FYI, there's also somewhat related AoPS thread of Somehow hard, but with the sum being equal to a fifth power. – John Omielan Aug 31 '23 at 20:24
  • @Hanno FYI, this answer of the proposed duplicate mentions that same result from Euler. Also, the other answers there generally give, either explicitly and/or with a link, additional results. – John Omielan Aug 31 '23 at 20:28
  • @JohnOmielan Do you know if there are infinitely many such numbers? – japjap Aug 31 '23 at 20:39
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    @japjap Based on only requiring the integers to be positive & distinct, there are infinitely many such numbers. Just take any solution, such as Euler's solution of $59^4+158^4=133^4+134^4$ (mentioned in Hanno's comment & mathlove's duplicate question answer), then multiply both sides by $k^4$ for any integer $k\gt 1$ to get $(59k)^4+(158k)^4=(133k)^4+(134k)^4$. However, apart from such relatively trivial results, i.e., with the additional condition of $\gcd(a,b,c,d)=1$, unfortunately I don't know much about it, including things like whether or not there are infinitely many such numbers. – John Omielan Aug 31 '23 at 20:50
  • Leo. Euler's instance (being minimal) is the generalised Taxicab number $T(4,2,2)$. There are more instances, also @JohnOmielan , cf
    https://mathworld.wolfram.com/DiophantineEquation4thPowers.html starting at $(115)$.     – Hanno Sep 01 '23 at 09:28
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    Still another one, quite large & not listed on the wolfraMathworld page $$4608535738441446497 = 1918^4 + 46333^4 = 35651^4 + 41594^4$$ – Hanno Sep 01 '23 at 09:49

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