Show that if x, y ∈ Z and 3|x(adsbygoogle = window.adsbygoogle || []).push({}); ^{2}+z^{2}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}= x^{2}+z^{2}+2xz

=> 3|x^{2}+z^{2}+2xz - (x^{2}+z^{2})

=> 3|2xz

=> 3|xz

so 3|x or 3|z

where to go from here?

just the arguement that if 3|x then 3|x^{2}

and therefore 3 must divide z^{2}=> 3|z?

Thanks

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# Homework Help: Another small number theory proof

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