3
$\begingroup$

This is what I have so far, I'm not sure my reasoning is correct as I am trying to learn how to construct proofs. I would appreciate any feedback on if I took the right steps. If there is an alternative way of going about this problem, what would it be? Thank you in advance.


def. of odd: $i = 2n + 1$

$i^4 = (2n + 1)^4 = (4n^2+4n+1)(4n^2+4n+1)$

From a previous problem, I showed that $4n^2+4n = 8j$. The problem was to show that the square of every odd number is of the form $8k+1.$

$i^4 = (8j+1)(8j+1) = 64j^2+14j+1 = 16j(4j+1)+1$

Here $16j$ is even and $4j+1$ is odd, so multiplying them would yield an even number.

I took a step back and substituted $64j^2+16j=16k$.

Therefore, $i^4 = (4n^2+4n+1) = (8j+1)(8j+1) = 16k+1$.

$\endgroup$
  • 1
    $\begingroup$ Even simpler, once you have $i^4=16j(4j+1)+1$, you are done, by taking $k=j(4j+1)$. $\endgroup$ – vadim123 Jul 10 '16 at 3:54
  • 1
    $\begingroup$ Your solution is correct, though the statement "Here $16j$ is.. even number" adds nothing to the proof $\endgroup$ – user258700 Jul 10 '16 at 3:54
  • $\begingroup$ @vadim123 I thought I wasn't allowed to assume that. $\endgroup$ – Curious Math Student Jul 10 '16 at 4:04
  • $\begingroup$ @AhmedHussein I see, thank you! $\endgroup$ – Curious Math Student Jul 10 '16 at 4:06
  • $\begingroup$ @vadim123 I forgot to thank you, sir! $\endgroup$ – Curious Math Student Jul 10 '16 at 4:06
0
$\begingroup$

$$(2k+1)^4-1 = 2k(2k+2)(4k^2+4k+2) = 2^4 \binom{k+1}{2}(2k^2+2k+1) \in 16\mathbb{Z}.$$

$\endgroup$
0
$\begingroup$

Here's a stupid way to do this question. $G = (\mathbb{Z}/16\mathbb{Z})^{\ast}$ is a group with $\varphi(16) = 2^4 - 2^3 = 8$ elements. The order of any element of $G$ is either $1, 2, 4$ or $8$ by Lagrange's theorem. So the only way the thing you are trying to prove could be false is if $G$ were cyclic.

If $G$ were cyclic, there would exist elements of order $8$. By a standard result about cyclic groups, number of such elements would be $\varphi(8) = 2^3 - 2^2 = 4$. This should leave exactly four elements with order $1, 2$ or $4$. You get a contradiction by checking that what you want to prove is true for at least five elements:

$$1^1 \equiv 1 \pmod{16}$$

$$3^4 = 81 \equiv 1 $$

$$5^4 = 625 = 39\cdot 16 + 1 \equiv 1$$

$$ 13^4 \equiv (-3)^4 \equiv 1$$

$$ 15^2 \equiv (-1)^2 \equiv 1$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.