anonymous
  • anonymous
Suppose a is an integer. Prove that if 32 ∤ ((a^2 + 3)(a^2 + 7)), then a is even.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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myininaya
  • myininaya
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
myininaya
  • myininaya
\[32m=(4k^2+4k+1+3)(4k^2+4k+1+7)\] type-o above
myininaya
  • myininaya
\[32m=(4k^2+4k+4)(4k^2+4k+8)\]

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myininaya
  • myininaya
\[32m=16k^4+16k^3+32k^2+16k^3+16k^2+32k+16k^2+16k+32\] \[32m=16k^4+k^3(16+16)+k^2(32+16+16)+k(32+16)+32\] \[32m=16k^4+32k^3+64k^2+48k+32\] \[2m=k^4+2k^3+4k^2+3k+2\] Thinking...
myininaya
  • myininaya
So somehow we need to show \[k^4+2k^3+4k^2+3k+2\] is even for any integer
myininaya
  • myininaya
We could try induction!
anonymous
  • anonymous
32m=(4k^2+4k+4)(4k^2+4k+8) 32m=16(k^2+k+1)(k^2+k+2)
anonymous
  • anonymous
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
myininaya
  • myininaya
you are right pizza! :)
myininaya
  • myininaya
great job!
anonymous
  • anonymous
Thanks! :)
anonymous
  • anonymous
np, myininaya did most of the work
anonymous
  • anonymous
:)
myininaya
  • myininaya
pizza i seen we had the same idea to do contrapositive lol
myininaya
  • myininaya
and you deleted
myininaya
  • myininaya
i don't think i could see a way to prove the statement as is
anonymous
  • anonymous
didn't know where to go with it, but the numbers looked too nice when you got to the 32m part
myininaya
  • myininaya
i was winging it
myininaya
  • myininaya
sometimes i don't know when i will run into a dead end
myininaya
  • myininaya
i just have to try
anonymous
  • anonymous
Anyways, Happy Birthday, XD
myininaya
  • myininaya
thanks :)

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