Question

How would I prove that if n is an integer and n^2 is divisible by 4, then n is even? I'm thinking about using induction, but not sure how that would work because (n+1)^2 isn't divisible by 4 if n^2 is, I think.

Answer 1

Plus, I don't have a smallest number in the set, so I'm not sure how I would prove a basis. I was thinking of maybe using induction to show it to be true for all n greater or equal to 0 and then showing that if it is true for that then it is true for all n <0 through direct proof, since (-n)^2=n^2. But not sure where to start

Answer 2

Any ideas? Maybe a hint? I appreciate it.

Answer 3

you must think... what are the possible squares?

1
4
9
16
25
36
...

4^2 = 16
6^2 = 36
8^2 = 64
10^2 = 100

etc. etc.

it does not matter for positive or negative - all numbers squared are positive

so, if the (n^2)/4 holds true, then n must have been even

because any odd value of n will make (n^2)/4 not true. notice the squares listed above, and it goes every other one (at least, for what i've written) for squares divisible by 4

Answer 4

hm, the intuition makes sense. I just need to figure out how to formalize it. Thanks for the help, it's a good place to start.

Answer 5

another way to look at it is all odd values squared will still be odd.

for instance
1^2 = 1
3^2 = 9
5^2 = 25
7^2 = 49

so, you could prove by contradiction and attempt to show it this way that any odd valued n will not be divisible by 4 when squared

Answer 6

it says nothing about all n so induction is not needed

Answer 7

Yeah, just tried doing proof by contradiction:

Suppose there exists an integer n such that n is odd and n^2 is divisible by 4. Then n=2k+1 for some integer k by the definition of odd and n^2=(2k+1)^2=4k^2+4k+1.

This factors to 4(k^2+k)+1, where k^2+1 is an integer since it's the sum of products of integers. But then n^2 is 4 times an integer plus 1, and is not divisible by 4.

Thanks for the help, I was banging my head against a wall there :)

Answer 8

@zzr0ck3r yeah, I was in a rut with the induction stuff

Answer 9

contradiction works, and contrposition would work also

Answer 10

There is a proof that says, if n is even, then n^2 is also even. (this proof also works for odd). so first set n to be even.
n=2k
Then plug that in for n^2. show that it's still divisible by 4.

Answer 11

your proof looks fine

Answer 12

hey thanks guys for the help, I really appreciate it.