show no solutions exists using table x^3mod9

Hi wondering if anyone can help with this question:

Make a table of values x^3, mod 9, and show that the equation x^2 + 2y^2 = 9z^2 has no non zero integer solutions.

My table looks like

n mod 9 - -4 -3 -2 -1 0 1 2 3 4

n^3mod 9 - -1 0 1 -1 0 1 -1 0 1

2n^3mod 9 --2 0 2 -2 0 2 -2 0 2

i can see that by adding the two rows i get 0's at 0 and +-3 but dont really know where to go from here.

Also if i am not given the modulo to use, how do i decide which modulo to use.

Thanks in advance

Re: show no solutions exists using table x^3mod9

Quote:

Originally Posted by

**MattWT** Hi wondering if anyone can help with this question:

My table looks like

n mod 9 - -4 -3 -2 -1 0 1 2 3 4 **?**

n^3mod 9 - -1 0 1 -1 0 1 -1 0 1 **?**

$\displaystyle n\ mod\ 9$ | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |

$\displaystyle n^3\ mod\ 9$ | 0 | 1 | 8 | 0 | 1 | 8 | 0 | 1 | 8 |

btw, are you sure you're not being asked to construct table for $\displaystyle x^2\ mod\ 9$ not $\displaystyle x^3\ mod\ 9$ ?

Re: show no solutions exists using table x^3mod9

You'll want to make a table for $\displaystyle x^2$ mod 9. For example,

$\displaystyle 1^2 \equiv 1$

$\displaystyle 2^2 \equiv 4$

$\displaystyle 3^2 \equiv 0$

$\displaystyle 4^2 \equiv 7$

$\displaystyle 5^2 \equiv 7$

$\displaystyle 6^2 \equiv 0$

$\displaystyle 7^2 \equiv 4$

$\displaystyle 8^2 \equiv 1$

$\displaystyle 9^2 \equiv 0$

Assume that at least one of x,y,z is not a multiple of 3. Even then, you can't prove that using the table, for example, you could have

$\displaystyle 1^2 + 2*7^2 \equiv 0 (mod 9)$