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PhoenixFire

  • 3 years ago

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

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  1. PhoenixFire
    • 3 years ago
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    For a given integer.

  2. carl51
    • 3 years ago
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    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

  3. zzr0ck3r
    • 3 years ago
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    hehe I wish, sec

  4. zzr0ck3r
    • 3 years ago
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    I would do contradiction

  5. PhoenixFire
    • 3 years ago
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    \[\forall{n}\in \mathbb{Z} : 4|n-2 \rightarrow 16|n^2-4\] I believe that's the correct notation.

  6. zzr0ck3r
    • 3 years ago
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    yeah

  7. zzr0ck3r
    • 3 years ago
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    do you need to show for all n?

  8. zzr0ck3r
    • 3 years ago
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    nm i c

  9. PhoenixFire
    • 3 years ago
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    I need to show the proof.

  10. zzr0ck3r
    • 3 years ago
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    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

  11. zzr0ck3r
    • 3 years ago
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    sorry direct proof was fast I think

  12. PhoenixFire
    • 3 years ago
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    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.

  13. zzr0ck3r
    • 3 years ago
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    yeah 16 | (16* any integer)

  14. PhoenixFire
    • 3 years ago
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    Thanks for the help.

  15. zzr0ck3r
    • 3 years ago
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    np

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