Can 201822018^{2} be represented as a3+b3+c3+d3a^{3}+b^{3}+c^{3}+d^{3}? Herea,b,c,dZa,b,c,d\in Z

I haven't find a concrete supportive example nor a proof of non-existence of such a representation yet.

But there are still some clues :

……………………………………………………………………………………………………………

First, realize 201824(mod9)2018^{2}\equiv 4 (mod9),

and xZ,x30,±1(mod9)\forall x\in Z,x^{3}\equiv 0,\pm 1(mod9) .

So if 201822018^{2} can be represented as a3+b3+c3+d3a^{3}+b^{3}+c^{3}+d^{3} for some a,b,c,dZa,b,c,d\in Z,

we certainly get a3b3c3d31(mod9)a^{3}\equiv b^{3}\equiv c^{3}\equiv d^{3}\equiv 1(mod9).

And further, a,b,c,d{9k+1kZ}{9k+4kZ}{9k+7kZ}a,b,c,d\in \{9k+1|k\in Z\}\cup\{9k+4|k\in Z\}\cup\{9k+7|k\in Z\}. ………………………………………………………………………………………………………………

_' Can,or cannot,this is a question! '

Note by Haosen Chen
1 year, 8 months ago

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This is still an open question. See (2) and (3) here.

Pi Han Goh - 1 year, 8 months ago

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Thank you,sir Pi Han Goh. My note was inspired by a problem in 2002 IMO proposal

━ Find the smallest positive integer n satisfying : the diophantine equation x13+x23+...+xn3=20022002x_{1}^{3}+x_{2}^{3}+...+x_{n}^{3}=2002^{2002} have integer solution ,

whose answer is 4.

Haosen Chen - 1 year, 8 months ago

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Ah good to know.

Originally, I spent over an hours trying to come up with a proof for your question that no solution exists via cubic residues but I failed badly so I decided to look it up.

Pi Han Goh - 1 year, 8 months ago

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@Pi Han Goh Oh,I'm sorry,sir. I said I was inspired because 20022002=(2002667)3(10003+10003+13+13)2002^{2002}=(2002^{667})^{3}\cdot (1000^{3}+1000^{3}+1^{3}+1^{3}) and then I wonder if 201822018^{2} can be done like 200212002^{1},since 20182(mod3)2018\equiv 2(mod3). So …

I'm sorry but you might fail to look it up.

Haosen Chen - 1 year, 8 months ago

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@Haosen Chen No, I totally got it. "I looked it up" = I give up because I don't know how to solve it, so I tried the internet for the answers.

Pi Han Goh - 1 year, 8 months ago

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