Thread: Prove that 2|(n^4 - 3) if and only if 4|(n^2+3)

1. Prove that 2|(n^4 - 3) if and only if 4|(n^2+3)

Course: Foundations of Higher Mathematics

Prove that $2|(n^4 - 3)$ if and only if $4|(n^2+3)$

I feel confident about this one. It's only the first part of the biconditional. Here's my attempt.

Assume that $2|(n^4-3)$ , i.e. $n^4-3=2a$, for some integer a.

Then, $n^4=2a+3$
$=2(a+1) +1$
$=2b+1$.
Since $a+1$ is an integer, b is an integer also.

Therefore $n^4$ is odd, which means that $n$ is odd, i.e. $n=2c+1$, for some integer c.

Then, $n^2+3=(2c+1)^2+3$
$=4c^2+4c+4$
$=4(c^2+c+1)=4d$, for some integer d

Therefore $n^2+3=4d$, hence $4|(n^2+3)$

Nicely done.

3. Re: Prove that 2|(n^4 - 3) if and only if 4|(n^2+3)

Prove that $2|(n^4 - 3)$ if and only if $4|(n^2+3)$
My take, which I don't know how to finish:
From the first relation: $n^4 - 3 \equiv 0 \text{ (mod 2)}$ which leads to $n \equiv 1 \text{ (mod 2)}$

And from the second: $n^2 + 3 \equiv 0 \text{ (mod 4)}$ which leads to $n \equiv {1, 3} \text{ (mod 4)}$

How do I finish? Is it the Chinese remainder theorem?

-Dan

4. Re: Prove that 2|(n^4 - 3) if and only if 4|(n^2+3)

Originally Posted by topsquark
My take, which I don't know how to finish:
From the first relation: $n^4 - 3 \equiv 0 \text{ (mod 2)}$ which leads to $n \equiv 1 \text{ (mod 2)}$

And from the second: $n^2 + 3 \equiv 0 \text{ (mod 4)}$ which leads to $n \equiv {1, 3} \text{ (mod 4)}$

How do I finish? Is it the Chinese remainder theorem?

-Dan
I'm not too familiar with modulo just yet. And I've never heard of that theorem before.

5. Re: Prove that 2|(n^4 - 3) if and only if 4|(n^2+3)

Originally Posted by topsquark
From the first relation: $n^4 - 3 \equiv 0 \text{ (mod 2)}$ which leads to $n \equiv 1 \text{ (mod 2)}$

And from the second: $n^2 + 3 \equiv 0 \text{ (mod 4)}$ which leads to $n \equiv {1, 3} \text{ (mod 4)}$
I am not sure which fact stated purely in terms of ≡ to use, but it is clear that n ≡ 1, 3 (mod 4) iff n is odd.

6. Re: Prove that 2|(n^4 - 3) if and only if 4|(n^2+3)

Originally Posted by emakarov
I am not sure which fact stated purely in terms of ≡ to use, but it is clear that n ≡ 1, 3 (mod 4) iff n is odd.
I can't believe I missed that. Thanks!

-Dan