3

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

Curious Math Student
  • 411

2 Answers2

1

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

Jack D'Aurizio
  • 353,855
0

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:

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

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

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

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

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

D_S
  • 33,891