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Prove that if n-2 is divisible by 4 then n^2 - 4 is divisible by 16.

Mathematics
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For a given integer.
Say that n is 26 so 26-2=24 and 24 is divisible by 4. 26x 2-4=52-4=48 and 48 is divisible by 16
hehe I wish, sec

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Other answers:

I would do contradiction
\[\forall{n}\in \mathbb{Z} : 4|n-2 \rightarrow 16|n^2-4\] I believe that's the correct notation.
yeah
do you need to show for all n?
nm i c
I need to show the proof.
ok assume n-2= 4k for some k in Z then n = 4k+2 then n^2-4 = (4k+2)^2 - 4 = 16k^2 + 16k +4-4 = 16(k^2+k) since k^2+k is in Z 16|n^2-4
sorry direct proof was fast I think
Yeah, they wanted Direct Proof. so since n^2 - 4 = 16k the (k^2+k) in 16(k^2+k) doesn't matter, the rest match. that's what was confusing me.
yeah 16 | (16* any integer)
Thanks for the help.
np

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