# Can $$2018^{2}$$ be represented as $$a^{3}+b^{3}+c^{3}+d^{3}$$? Here$$a,b,c,d\in Z$$

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 $$2018^{2}\equiv 4 (mod9)$$,

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

So if $$2018^{2}$$ can be represented as $$a^{3}+b^{3}+c^{3}+d^{3}$$ for some $$a,b,c,d\in Z$$,

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

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

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

Note by Haosen Chen
4 months ago

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

- 4 months ago

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 $$x_{1}^{3}+x_{2}^{3}+...+x_{n}^{3}=2002^{2002}$$ have integer solution ,

- 4 months ago

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.

- 4 months ago

Oh,I'm sorry,sir. I said I was inspired because $$2002^{2002}=(2002^{667})^{3}\cdot (1000^{3}+1000^{3}+1^{3}+1^{3})$$ and then I wonder if $$2018^{2}$$ can be done like $$2002^{1}$$,since $$2018\equiv 2(mod3)$$. So …

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

- 4 months ago

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.

- 4 months ago