# Another small number theory proof

1. May 9, 2010

### Firepanda

Show that if x, y ∈ Z and 3|x2+z2 then 3|x and 3|z

Solution:

3|(x-z)(x+z)

=> 3|x+z or 3|x-z

if 3|x+z

then 3|(x+z)2 = x2+z2 +2xz

=> 3|x2+z2 +2xz - (x2+z2)

=> 3|2xz

=> 3|xz

so 3|x or 3|z

where to go from here?

just the arguement that if 3|x then 3|x2

and therefore 3 must divide z2 => 3|z?

Thanks

Last edited: May 9, 2010
2. May 9, 2010

### Office_Shredder

Staff Emeritus
That's pretty much it

3. May 9, 2010

### Petek

Is the statement of the problem correct? We see a "y" in the hypothesis, but a "z" in the conclusion. More serious, instead of

3|x2+z2

did you mean

3|x2-z2?

I'm guessing that you meant the latter since your solution starts with

3|(x-z)(x+z)

and (x-z)(x+z) = x2-z2

Finally, you had two cases (3|x+z or 3|x-z), but only addressed the first.

Petek

4. May 9, 2010

### Dick

It can't be x^2-z^2. If x=4 and z=2, 3|12, but 3 doesn't divide 4 OR 2. Firepanda should try thinking about what x^2 mod 3 equals.

Last edited: May 9, 2010