Hello guys!

We are to prove that if 7 divides $\displaystyle x^3+y^3+z^3$, then it also divides $\displaystyle xyz$. How do we go about it?

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- Oct 24th 2012, 06:11 AMcosmonavtProve if 7 divides x^3+y^3+z^3 then it also divides xyz
Hello guys!

We are to prove that if 7 divides $\displaystyle x^3+y^3+z^3$, then it also divides $\displaystyle xyz$. How do we go about it? - Oct 24th 2012, 07:56 AMDavidErikssonRe: Prove if 7 divides x^3+y^3+z^3 then it also divides xyz
Prove that $\displaystyle a^3 \mbox{mod}(7)$ can only take the values 0,1,6 by considering $\displaystyle a=7n+k$ for $\displaystyle k=0,1,..,6$. After this it follows at once that at least one of x,y or z must be divisible by 7.

- Oct 24th 2012, 08:27 AMcosmonavtRe: Prove if 7 divides x^3+y^3+z^3 then it also divides xyz
I don't have much experience with modular congruences. Can you please elaborate?

- Oct 24th 2012, 10:41 AMDavidErikssonRe: Prove if 7 divides x^3+y^3+z^3 then it also divides xyz
Modulo is the rest after division so what I want you to show is that no matter what integer you have, if you take it to the power of three and after that divide by seven the rest must be either 0, 1 or 6. The main idea is that if you know what the rest is for say a=1 is then you know that it must be the same for a=8=7+1 since you can expand using the binomial theorem and all terms will contain a 7 except for 1^3 and so on. That means that you plug in 0,1,...,6 and check what the rest is after division with 7 and from the possible rests you can see that one of them has to be 0 and that proves the statement.

- Oct 24th 2012, 10:45 PMSalahuddin559Re: Prove if 7 divides x^3+y^3+z^3 then it also divides xyz
In fact, this is true for any odd number of cubes summed.

Salahuddin - Oct 24th 2012, 11:32 PMDavidErikssonRe: Prove if 7 divides x^3+y^3+z^3 then it also divides xyz
That doesn't hold for say 7 cubes

- Oct 25th 2012, 05:19 AMSalahuddin559Re: Prove if 7 divides x^3+y^3+z^3 then it also divides xyz
Yes, thats right, I missed that things. Whenever the number of cubes is n, and there exists an x, such that (0 <= x <= n), (n + 5x) mod 7 = 0, this does not work. I understand that this is a much more complex formalization of it. By the same argument, there exists such an x, for all (n >= 7). So, in fact, it works only for 1, 3, and 5 where it holds for 1 trivially.

Salahuddin

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