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
7 months, 1 week ago

No vote yet
1 vote

  Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 \( 2 \times 3 \)
2^{34} \( 2^{34} \)
a_{i-1} \( a_{i-1} \)
\frac{2}{3} \( \frac{2}{3} \)
\sqrt{2} \( \sqrt{2} \)
\sum_{i=1}^3 \( \sum_{i=1}^3 \)
\sin \theta \( \sin \theta \)
\boxed{123} \( \boxed{123} \)

Comments

Sort by:

Top Newest

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

Pi Han Goh - 7 months, 1 week ago

Log in to reply

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 ,

whose answer is 4.

Haosen Chen - 7 months, 1 week ago

Log in to reply

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 - 7 months, 1 week ago

Log in to reply

@Pi Han Goh 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.

Haosen Chen - 7 months, 1 week ago

Log in to reply

@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 - 7 months, 1 week ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...