Given the equation:
$$ x^4 + y^4 = k, $$
where $x$, $y$ and $k$ are distinct non-zero integers, is there any $k$, such that there is more than one solution $\{x, y\}$ for the above equation?
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1see https://en.wikipedia.org/wiki/Generalized_taxicab_number – mathlove Aug 22 '15 at 13:10
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@mathlove Thanks for that link above. I am a complete no-nothing in number theory so that would serve as a starting point. – balajeerc Aug 22 '15 at 13:12
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@mathlove Following the references in the page you linked to above, it appears there is at least one solution: $$ 1584^4 + 594^4 = 1344^4 + 1334^4 = 635318657 $$ (Euler, 1772). Would you like to convert your question to an answer so I can accept it? – balajeerc Aug 22 '15 at 13:15
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You have put a superfluous 4 as the last digit of each of your numbers, bala. – Gerry Myerson Feb 03 '19 at 06:48
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Should Be Editted. Should say "4th powers" instead of "powers of 4"! – Mike Sep 02 '21 at 00:44
5 Answers
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Euler showed that $$635318657 = 59^4+158^4 =133^4 + 134^4$$ is the smallest number which can be expressed as the sum of $2$ $4$th positive powers in $2$ different ways.
mathlove
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The biggest number that I found can be expressed as a sum of two 4th powers in more than 1 way is: $2602265219072= 1064^4 + 1072^4=472^4+1264^4$ Guys, guess what, I found the other way to represent 11220039255312 as a sum of two 4th powers in 2 different ways:
$11220039255312=1752^4+1158^4= 1536^4+1542^4$
I bet if anyone can find a bigger number than 11220039255312 which can be represented as a sum of two 4th powers in two different ways
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2This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review – Oshawott Sep 01 '21 at 14:14
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2@Unknown Well, it does answer the question. The question asks if there exists some $k$ which can be expressed as a sum of two 4th powers in more than 1 way. This answer says $k=2602265219072$ works. (One of the other answers gives a link with lots of other solutions, and the solution here is bigger than all of them.) – user1729 Sep 01 '21 at 14:47
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1@Unknown, please check properly, it does work and it answers the question also – SHOUNAK GUPTA Sep 01 '21 at 14:48
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I found an even bigger number , that is,11220039255312 which can be expressed as follows: 11220039255312=1536⁴ +1542⁴ But unfortunately, I cannot find the way to represent it in another way but my computer says that it can be represented in 2 ways – SHOUNAK GUPTA Sep 01 '21 at 14:50
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So, can anyone find another way to represent 11220039255312 as a sum of two positive 4th powers in another way? – SHOUNAK GUPTA Sep 01 '21 at 15:13
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1@SHOUNAK You could ask this as a new question, and link back to this one. – user1729 Sep 03 '21 at 07:06
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Yeah.... it's a good idea....I might do that but did u see my second answer? It's even bigger than this one!! – SHOUNAK GUPTA Sep 03 '21 at 12:03
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This is just a generalization of the famous Ramanujan taxicab problem. We are looking for integer solutions of $$ x^4+y^4 = z^4+w^4 $$ or: $$ x^4-z^4 = w^4-y^4 \tag{1}$$ with $\{x,y\}\neq\{z,w\}$. Euler found:
$$635318657 = 133^4 + 134^4 = 158^4 + 59^4\tag{2}$$
and that is the smallest solution.
Jack D'Aurizio
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Here is a list of results by Jarek Wroblewski http://www.math.uni.wroc.pl/~jwr/422/422-10m.txt, the first of which is the famous one by Euler already cited
David Quinn
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Since, nobody could find a greater number than 11220039255312 , so, I found it on my own and here it is: $26033514998417=2189^4+1324^4=1784^4+1997^4$
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This one also works, and this is the biggest solution I've found yet. – SHOUNAK GUPTA Sep 03 '21 at 07:41