# Solution

Show that the square of every odd integer is of the form $8n+1$ and the fourth power of every odd integer is of the form $16k+1$

Part I.
Let $a$ be an odd integer. Then $a=2k_1+1$ for some integer $k_1$. Consider the square of $a$: \begin{align}a^2 &= (2k_1 + 1)^2\\ &= 4{k_1}^2 + 4k_1 + 1\\ &= 4({k_1}^2 + k_1) + 1\end{align} So, certainly the square of an odd integer is of the form $4k+1$, but we hoped this to be of the form $8k+1$. If we can argue that $({k_1}^2 + k_1)$ must be even, that would give us the additional factor of two that we need.

Notice, $k_1$ must either be even or odd.

In the first case, when $k_1$ is even, notice that ${k_1}^2$ must also be even, making ${k_1}^2+k_1$ the sum of two even numbers, which must be even.

In the second case, when $k_1$ is odd, notice that ${k_1}^2$ must also be odd, making ${k_1}^2+k_1$ the sum of two odd numbers, which must be even.

In both cases, ${k_1}^2+k_1$ is even, and hence, we can find some integer $k_2$ so that ${k_1}^2+k_1 = 2k_2$.

Now, picking up where we left off above, we have: \begin{align}a^2 &= 4({k_1}^2 + k_1) + 1\\ &= 4(2k_2)+1\\ &= 8k_2 + 1\end{align} Thus, the square of every odd integer is of the form $8k+1$.

Part II.
Now consider the fourth power of this odd integer $a$. Note, we have already determined above that $a^2 =8k_2+1$ for some integer $k_2$. So, \begin{align}a^4 &= (a^2)^2\\ &= (8k_2 + 1)^2\\ &= 64{k_2}^2 + 16k_2 + 1\\ &= 16(4{k_2}^2 + k_2) + 1\end{align} We know that $(4{k_2}^2 + k_2)$ must be an integer due to the closure of the integers under addition and multiplication. So, defining $k_3 = 4{k_2}^2 + k_2$, we see that $a^4$ can be written as $16k_3 + 1$. Thus, the fourth power of every odd integer is of the form $16k+1$.