Modular arithmetic three variables

Question

Show that if the integers $x, y,$ and $z$ satisfy $x^3 + 3y^3 = 9z^3$ then $x = y = z = 0.$ How should I interpret this question and how to proceed? I am thinking about the Euclidean algorithm but it becomes confusing when $x,y,z$ comes like variables?

This is actually proved by infinite descent, not with modular arithmetic. — Crostul, Oct 28 '16 at 22:23

score 3 · Accepted Answer · edited Oct 28 '16 at 23:17

3

First notice that if $d=\mbox{gcd}(x,y,z)$ then $d^3$ can be factored out of the equation. So we can assume that $d=1$. Then $x^3 = 9z^3-3y^3$, so $3$ divides $x$, say $ x=3k$. So we have $3^3k^3 = 9z^3-3y^3$ and we can divide everything by $3$ to get $9k^3= 3 z^3-y^3$. A similar argument shows $3$ divides $y$. Repeat to show $3$ divides $z$. This contradicts that $d=1$.

edited Oct 28 '16 at 23:17

Ethan Bolker

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answered Oct 28 '16 at 22:31

B. Goddard

1

Are i testing just to divide with 3=?? – user3704516 Oct 28 '16 at 22:42
Yes. 3 is the key to the problem. – B. Goddard Oct 28 '16 at 23:16

score 0 · Answer 2 · answered Oct 28 '16 at 23:15

0

Hint $\ $ The exponent of $\,3\,$ on the RHS is $\equiv 2\pmod 3\,$ but is $\equiv 0$ or $1\pmod 3$ on the LHS.

answered Oct 28 '16 at 23:15

Bill Dubuque

272,048

Modular arithmetic three variables

2 Answers2