Question

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

Answer 1

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

Answer 2

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

Answer 3

\[32m=(4k^2+4k+4)(4k^2+4k+8)\]

Answer 4

\[32m=16k^4+16k^3+32k^2+16k^3+16k^2+32k+16k^2+16k+32\]
\[32m=16k^4+k^3(16+16)+k^2(32+16+16)+k(32+16)+32\]
\[32m=16k^4+32k^3+64k^2+48k+32\]
\[2m=k^4+2k^3+4k^2+3k+2\]

Thinking...

Answer 5

So somehow we need to show
\[k^4+2k^3+4k^2+3k+2\] is even for any integer

Answer 6

We could try induction!

Answer 7

32m=(4k^2+4k+4)(4k^2+4k+8)
32m=16(k^2+k+1)(k^2+k+2)

Answer 8

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

Answer 9

you are right pizza! :)

Answer 10

great job!

Answer 11

Thanks! :)

Answer 12

np, myininaya did most of the work

Answer 13

:)

Answer 14

pizza i seen we had the same idea to do contrapositive lol

Answer 15

and you deleted

Answer 16

i don't think i could see a way to prove the statement as is

Answer 17

didn't know where to go with it, but the numbers looked too nice when you got to the 32m part

Answer 18

i was winging it

Answer 19

sometimes i don't know when i will run into a dead end

Answer 20

i just have to try

Answer 21

Anyways, Happy Birthday, XD

Answer 22

thanks :)