Numbers greater than $26033514998417$ that can be expressed as a sum of two positive 4th powers in more than 1 way

Question

Are there any numbers greater than $26033514998417$ which can be expressed as a sum of two positive 4th powers in atleast 2 ways ?

And if there are any such numbers, then, please include the number(s) that u have found in your answer.

In case, if anybody wants to know how the number

$26033514998417$ can be expressed as a sum of two positive 4th powers in 2 ways, here it is: https://math.stackexchange.com/a/4237040/857041

The conditions must be as follows: x^4+y^4=k, where x,y and k are non-zero integers , such that there is more than 1 solution{x,y} for the above equation and k>$26033514998417$ — SHOUNAK GUPTA, Sep 17 '21 at 20:44
You mean, apart from multiplying the given number by 16, 81, etc? — Oscar Lanzi, Sep 17 '21 at 20:45

emacs drives me nuts · Accepted Answer · 2021-09-18T14:04:58.480

3

$$17472238301875630082 = 62047^4 + 40351^4 = 59693^4 + 46747^4$$ from A018786, all primes btw.

Dunno why that list stops at 835279626752, though, presumably because a sequence must be complete, i.e. all numbers smaller and not in the list must not have the property.

Addendum: and here is a list with 516 primitive entries from Daniel J. Bernstein (found via A003824), that ends

...
955378 168531 883693 688026
960319  87924 942337 498984
962724 455837 870557 756684
972841 283961 967979 394717
988582 158157 976563 463702
989216 283949 958163 590024
989426 503473 938698 704399
989727 161299 913141 717447
990518   1039 967823 540326

Addendum 2: Here is a 146-digit solution from PrimePuzzles 103

75326517042882955509049316560407015204559148957492168402274838923500008575948738239692316743036018051328508933923770491774189595934532956237348642
= 1679539956802461427023806692932554869^4
+ 2864939822128245005298014063613916133^4
=  735636962517662175684582548040182073^4
+ 2096549864621014042130013452441703601^4

They mention an identity from a collection by Edward Brisse on EulerNet: $$f_2^4(a,b)+f_2^4(b,-a)=f_2^4(a,-b)+f_2^4(b,a)$$ where $$f_2(a,b)= -a^{13}+a^{12}b+a^{11}b^2+5a^{10}b^3+6a^9b^4-12a^8b^5-4a^7b^6+7a^6b^7-3a^5b^8-3a^4b^9+4a^3b^{10}+2a^2b^{11}-ab^{12}+b^{13}$$

The collection has also (other) identidies of degree 7, 13, 19 and 31 for $a$, $b$, $c$, $d$. And there are identities that solve $x^4+y^4+z^4=2w^4$ like

$$(a^2-b^2)^4 + (a^2 + 2ab)^4 + (2ab + b^2)^4 = 2(a^2 + ab + b^2)^4$$

So this rabbit hole seems to go infinitly deep, both in terms of formulae and in terms of solutions you can generate with each formula.

edited Sep 18 '21 at 14:04

answered Sep 17 '21 at 20:57

emacs drives me nuts

10,390
2
12
31

Thank you for the answer – SHOUNAK GUPTA Sep 17 '21 at 21:00
1

@SHOUNAK GUPTA: I found a list with even bigger primitive entries, see "Addendum". – emacs drives me nuts Sep 17 '21 at 21:12
1

$962608047985759418078417 = 990518^4+1039^4=967823^4+540326^4$ has 24 digits. – emacs drives me nuts Sep 17 '21 at 21:25
Yes, u are correct....I also saw a list from D.Wilson:https://oeis.org/A003824/b003824.txt – SHOUNAK GUPTA Sep 17 '21 at 21:30
1

The largest number comes from the $514^{th}$ instead of $516^{th}$ entry. It is $1022624816783516967864017 = 989426^4 + 503473^4 = 938698^4 + 704399^4$ – achille hui Sep 17 '21 at 21:51
And the paper where Daniel J.Bernstein talks about "Enumerating solutions to p(a)+q(b)=r(c)+s(d)" is here:. Here's the link to the file:
https://in.docworkspace.com/d/sALIKc77niZk6q5yMiaqnFA

Shared from WPS Office: https://kso.page.link/wps
– SHOUNAK GUPTA Sep 17 '21 at 21:51
1

@SHOUNAK GUPTA: See "Addendum 2" for bigger solutions and generic formulae to generate solutions. – emacs drives me nuts Sep 18 '21 at 09:01
Yeah...They really do go infinitely deep.... Anyways, thank you so much for spending your precious time in answering my question and a special thanks for the"Addendum 2" :) – SHOUNAK GUPTA Sep 18 '21 at 11:16

Numbers greater than $26033514998417$ that can be expressed as a sum of two positive 4th powers in more than 1 way

1 Answers1