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I want to prove that $\mathrm{gcd}(x-4,x+4)$ divides $8$ for all $x\in \mathbb{Z}$

Since they are both polynomials of degree $1$, it suggests that the $\mathrm{gcd}$ is a constant.

Using Euclidean Algorithm, I get: $(x+4) = 1(x-4) + 8$, so $\mathrm{gcd}(x-4,x+4)=\mathrm{gcd}(x-4,8)$ thus it will always divide $8$.

Is this the correct approach / use of EA for polynomials?

Thanks.

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  • $\begingroup$ Yes, but for integers, not polynomials. See here for more. $\endgroup$ – Bill Dubuque Jan 30 '15 at 18:00
  • $\begingroup$ Ah okay. So this approach works solely because I am evaluating x-4 and x+4 at some integer x, so it is essentially just using EA for integers? $\endgroup$ – lifin Jan 30 '15 at 18:07
  • $\begingroup$ The gcd is not a gcd of polynomials, but integers, e.g. $\,\gcd(x,2) = 1\,$ in $\,\Bbb Z[x]\,$ but the gcd $= 2$ if $\,x\,$ is an even integer. You can deduce some integer information from polynomial gcds, but there is no need to even consider polynomials here - just work with integers, i.e. use $\,\gcd(a,b) = \gcd(a\!-\!b,b)\,$ for all integers $\,a,b.\ \ $ $\endgroup$ – Bill Dubuque Jan 30 '15 at 18:13
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Your approach is correct. Another way of proving it is the following: $$ d\mid x-4\text{ and }d\mid x+4\implies d\mid(x+4)-(x-4)=8. $$

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  • $\begingroup$ Thanks, I just thought about your approach now as well and was about to ask, so good timing! $\endgroup$ – lifin Jan 30 '15 at 17:51
  • $\begingroup$ Actually, I went with $d|(x+4)-(x-4) \Rightarrow d|8$. Is that what you meant to write? $\endgroup$ – lifin Jan 30 '15 at 17:52
  • $\begingroup$ Yes, that is what I meant. I will edit my answer. $\endgroup$ – Julián Aguirre Jan 30 '15 at 17:53

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