# Congruency Question

• Sep 15th 2010, 07:52 PM
Flippinpony
Congruency Question
Hey, first post and all so maybe someone can help me out with a question from my Number Theory class. It's in the Congruences chapter.

Prove that if n is congruent to 4 (mod 9), then n cannot be written as the sum of 3 cubes.

Thanks in advance for the help!
• Sep 15th 2010, 07:58 PM
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Quote:

Originally Posted by Flippinpony
Hey, first post and all so maybe someone can help me out with a question from my Number Theory class. It's in the Congruences chapter.

Prove that if n is congruent to 4 (mod 9), then n cannot be written as the sum of 3 cubes.

Thanks in advance for the help!

$\displaystyle 0^3\equiv0\pmod{9}$

$\displaystyle 1^3\equiv1\pmod{9}$

$\displaystyle 2^3\equiv8\pmod{9}$

$\displaystyle 3^3\equiv0\pmod{9}$

$\displaystyle 4^3\equiv1\pmod{9}$

$\displaystyle 5^3\equiv8\pmod{9}$

$\displaystyle 6^3\equiv0\pmod{9}$

$\displaystyle 7^3\equiv1\pmod{9}$

$\displaystyle 8^3\equiv8\pmod{9}$

See where to go from here?
• Sep 15th 2010, 08:17 PM
Flippinpony
Ok, so what I'm seeing is that all cubes are congruent to either 0, 1, or 8 (mod 9). Therefore, a cube or a sum of cubes for that matter cannot be congruent to 4 (mod 9). Is there a theorem I should point to when writing a proof for this?
• Sep 15th 2010, 08:25 PM
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Quote:

Originally Posted by Flippinpony
Ok, so what I'm seeing is that all cubes are congruent to either 0, 1, or 8 (mod 9). Therefore, a cube or a sum of cubes for that matter cannot be congruent to 4 (mod 9). Is there a theorem I should point to when writing a proof for this?

That's essentially all there is, it's just an exhaustive search. Maybe there's a more elegant way, I don't know. You can list systematically and for each easily verify the sum is not 4 mod 9.

000
001
008
011
018
088
111
118
188
888

Edit: I'm not sure if it's a typo when you wrote that a sum of cubes can't be congruent to 4 mod 9... of course if we have four cubes then we can just choose 1+1+1+1.