$n$ is odd if and only if there exists an $a \in \mathbb{Z}$ such that $n^4=16a+1$ Write a formal proof.
Suppose that $n \in ℕ$. Prove $n$ is odd if and only if there exists $a \in \mathbb{Z}$ such that $n^4 = 16a + 1$.
There exists $k\in \mathbb{Z}$ such that $n=2k+1$. So I've used this formula for odd numbers
$$n^4=(2k+1)^4 = 16k^4+32k^3+24k^2+8k+1$$
since our goal is to match this to the above $n^4$,
$$n^4=8(2k^4+4k^3+3k^2+k)+1$$
unfortunately this form does not match $16a+1$
This is where I got stuck and could use help on the proof.
A random solution I thought of is below, however I'm looking for a better answer.
Let $a=2k$ ; I randomly though of a substitution.
$$n^4=8(2*16a^4+16*8a^3+3*4a^2+2a)+1$$
$$n^4=16(16a^4+64a^3+6a^2+a)+1$$
this has the form $16a+1$ with '$(16a^4+64a^3+6a^2+a)$ = an integer'.
Note
Thanks everyone for the help the proof totally makes sense now!
Special shoutout to @lulu and @fleablood
The part that solidified it for me was the breakdown of the 2 cases
*(3+1)2, k being even or odd!
 A: If $n$ is odd then there exists an integer $k$ so that $n$ can be written as $n=2k +1$. And with respect to that $k, n^4 = (2k+1)^4 = 16k^4 + 4*8k^3 + 6*4k^2 +   4k + 1=16k^4 + 32k^2 + 24k^2 + 4k +1$.
And $n^4= 16(k^4 + 2k^3 + \frac {3k^2 + k}2) + 1= 16(k^4 + 2k^3 + \frac {k(3k+1)}2) + 1$.
So if we can prove $\frac {k(3k+1)}2$ is always an integer we are done.  If $k$ is even than $\frac k2$ is an integer. and $\frac k2(3k+1)$ is an integer.  If $k$ is odd then $3k+1$ is even so $\frac {3k+1}2$ is an integer.
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In hindsight we could have realized that if $n$ is odd then $n$ can be written as $4k \pm 1$ for some $k$ and some choose of $+/-$.
Then $n^4 = (4k \pm 1)^4 = 4^4k^4 \pm 4*4^3k^3 + 6*4^2k^2 \pm 4*4k + 1= 16(16k^4 \pm 16k^3 +6k^2 \pm k) + 1$.
And if we want to be pedantic masochists  $n^4 = 16a+1$ when $a = (\frac {n-1}2)^4 + 2(\frac {n-1}2)^3 + \frac {(n-1)(3n-1)}4$, which, as $n$ is odd is an integer.
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That proves if $n$ is odd then there is an integer $a$ so that $n^4 = 16a + 1$.
If $n^4 = 16a+1$ then $n^4$ is odd and $n$ is odd.  (So the other direction is facile.)
A: It's easier if done in two stages:
$n^2 = (2k+1)^2 = 4k^2+4k+1 = 8\binom{k+1}{2}+1 = 8b+1$
$n^4=(8b+1)^2= 64b^2+16b+1 = 16(4b^2+b)+1 = 16a+1$
