Another small number theory proof

[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 argument that if 3|x then 3|x²

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

Thanks

 

[Office_Shredder]

That's pretty much it

 

[Petek]

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 argument that if 3|x then 3|x²

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

Thanks

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

 

[Dick]

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

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.