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anonymous

  • 4 years ago

Suppose a is an integer. Prove that if 32 ∤ ((a^2 + 3)(a^2 + 7)), then a is even.

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  1. myininaya
    • 4 years ago
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    Let me take a stab at this. So I'm thinking to prove the contrapositive. If a is odd, then 32|((a^2+3)(a^2+7)). Let a=2k+1 since a is odd for some integer k. Now 32|((a^2+3)(a^2+7)) =>there is some integer m such that 32m=(a^2+3)(a^2+7) Now remember we let a=2k+1 So we have \[32m=((2k+1)^2+3)((2k+1)^2+7)\] \[32m=(4k^2+4k+1+3)(4k+4k+1+7)\] .... This is what I'm thinking so far Still working

  2. myininaya
    • 4 years ago
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    \[32m=(4k^2+4k+1+3)(4k^2+4k+1+7)\] type-o above

  3. myininaya
    • 4 years ago
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    \[32m=(4k^2+4k+4)(4k^2+4k+8)\]

  4. myininaya
    • 4 years ago
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    \[32m=16k^4+16k^3+32k^2+16k^3+16k^2+32k+16k^2+16k+32\] \[32m=16k^4+k^3(16+16)+k^2(32+16+16)+k(32+16)+32\] \[32m=16k^4+32k^3+64k^2+48k+32\] \[2m=k^4+2k^3+4k^2+3k+2\] Thinking...

  5. myininaya
    • 4 years ago
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    So somehow we need to show \[k^4+2k^3+4k^2+3k+2\] is even for any integer

  6. myininaya
    • 4 years ago
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    We could try induction!

  7. anonymous
    • 4 years ago
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    32m=(4k^2+4k+4)(4k^2+4k+8) 32m=16(k^2+k+1)(k^2+k+2)

  8. anonymous
    • 4 years ago
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    since one of (k^2+k+1) or (k^2+k+2) is even, thus 16(k^2+k+1)(k^2+k+2) divides 32m

  9. myininaya
    • 4 years ago
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    you are right pizza! :)

  10. myininaya
    • 4 years ago
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    great job!

  11. anonymous
    • 4 years ago
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    Thanks! :)

  12. anonymous
    • 4 years ago
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    np, myininaya did most of the work

  13. anonymous
    • 4 years ago
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    :)

  14. myininaya
    • 4 years ago
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    pizza i seen we had the same idea to do contrapositive lol

  15. myininaya
    • 4 years ago
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    and you deleted

  16. myininaya
    • 4 years ago
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    i don't think i could see a way to prove the statement as is

  17. anonymous
    • 4 years ago
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    didn't know where to go with it, but the numbers looked too nice when you got to the 32m part

  18. myininaya
    • 4 years ago
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    i was winging it

  19. myininaya
    • 4 years ago
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    sometimes i don't know when i will run into a dead end

  20. myininaya
    • 4 years ago
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    i just have to try

  21. anonymous
    • 4 years ago
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    Anyways, Happy Birthday, XD

  22. myininaya
    • 4 years ago
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    thanks :)

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