$\sum_{n=1}^{2017}\left(\left((n+2)^4\bmod{(n+1)^4}\right)\bmod{4}\right)$ 
What's
$$\sum_{n=1}^{2017}\left(\left((n+2)^4\bmod{(n+1)^4}\right)\bmod{4}\right)$$

What have I tried?
$$(n+2)^4=n^4+8n^3+24n^2+32n+16$$
$$(n+1)^4=n^4+4n^3+6n^2+4n+1$$
Remainder:
$$4n^3+18n^2+28n+15$$
mod:
$$2n^2-1\pmod{4}$$
I can compute $\sum x^2$ but I don't know what to do with $$\sum_{n=1}^{2017}\left(2n^2-1\mod{4}\right)$$
 A: Hint: Compute the sum of even and odd $n$ separately. That is, $$\sum_{n=1}^{2b+1}{(2n^2-1\mod{4})} = \sum_{k=1}^{b}{(2(2k)^2-1\mod{4})} + \sum_{k=0}^{b}{(2(2k+1)^2-1\mod{4})}$$
Just a word of caution: If $a \mod{4} \in \{0,1,2,3\}$ then it is useful to use $2n^2+3$ instead of $2n^2-1$, especially when $n$ is even.
A: 
Here
  $$\{\}\text{ is the fractional part function}$$
  So we can write, (if you have problem with this part please comment)
  $$\text{a mod b}= \{\frac{a}{b}\} b$$

$$\sum_{n=1}^{2017} \{\frac{2 n^2-1}{4}\} 4$$
$$\sum_{n=1}^{2017} \{\frac{ n^2}{2}-\frac{1}{4}\} 4$$

$$\{\frac{ n^2}{2}-\frac{1}{4}\}= \frac {1}{4}  \text { for } n\in odd$$
  $$ =\frac{3}{4} \text{ for }n \in even $$

$$\sum_{n=1}^{1009} \frac{1}{4}4 +\sum_{n=1}^{1008} \frac{3}{4} 4$$
$$\sum_{n=1}^{1009} 1 +\sum_{n=1}^{1008}  3 $$
$$1009 + 3 × 1008$$

$$4033$$

A: What a mess.
First of I must point out that the is meaningless garbage.
$a \mod b + c \mod d$ is an abuse of notation.  The statement $a \mod b$ is not a number.  It is a class of equivalent numbers.  It may but add moduloly to other classes in the same modulo residue system but not to classes in others.
However I will take this (under objection) as an abuse of notation for the remainder function where, for example $28 \mod 5 = 3$ and not $-2$ or $7$ as we are taking one value between $0$ and $5$.  This is wrong and you should use a different notation for the remainder function that returns a number.  
But I'll use your abuse.
======
But note:  $(k+1)^4 \mod k^4 \equiv 4k^3 + 6k^2 + 4k + 1 \mod k^4$
So $(n+2)^4 \mod (n+1)^4 \equiv 4(n+1)^3 + 6(n+1)^2 + 4(n+1) + 1 \mod (n+1)^4$
And modulo $4$ we get:
$[ (n+2)^4 \mod (n+1)^4] \mod 4 \equiv [2(n+1)^2 + 1]\mod 4 \equiv 2n^2 + 4n + 3\equiv 2n^2 + 3 \mod 4$.
If $n \equiv 0,1,2,3 \mod 4$ we have $2n^2 + 3 \equiv 3,1,3,1 \mod 4$.
So $\sum\limits_{i= 2k+1}^{2k+2} ([ (n+2)^4 \mod (n+1)^4] \mod 4) = 1+ 3 = 4$.
So  $\sum\limits_{i= 1}^{2017} ([ (n+2)^4 \mod (n+1)^4] \mod 4)=\sum\limits_{i=1}^{2017}[3\text{ if i is odd}|1\text{ if i is even}]= \frac {2016}2*4 + 3 = 4035$.
A: For odd $n$, $(n+1)^4$ is a multiple of $4$, so in that case $$(X\bmod(n+1)^4)\bmod 4 = X \bmod 4$$ and we just have to sum $(n+2)^4\bmod 4$. But $(n+2)^4$ is an odd square, and all odd squares are $1$ modulo $4$.
For even $n$ your calculation of $(n+2)^2 \bmod (n+1)^4 = 4n^3+18n^2+28n+15$ works as long as $(n+1)^4 < 2(n+1)^4$ which happens when $n>\frac{1}{\sqrt[4]2-1}-1 \approx 4.3$. So the cases $n=2,4$ need to be treated as special cases.
Otherwise you're summing (as you say) $2n^2-1$ modulo $4$. But since $n$ is now even, $2n^2$ will be a multiple of $4$, so it disappears and only the $-1$ is left.
So for $n\ge 4$ the $+1$ from an odd $n$ cancels out the $-1$ from the even $n$ next to it, and all that is left is
$$\sum_{n=1}^{3}\left(\left((n+2)^4\bmod{(n+1)^4}\right)\bmod{4}\right)$$
which can be done by hand easily.
A: For this term: $\left((n+2)^4\bmod{(n+1)^4}\right)$, it's easy to see that for $n\ge 5$, this is just $\left((n+2)^4-{(n+1)^4}\right)$, since $\frac{7^4}{6^4}<2$. So after the first four terms we can just use the pattern of values $\bmod 4$ to complete the calculation, since for $k$ odd, $k^4\equiv 1 \bmod 4$ and for $k$ even, $k^4\equiv 0 \bmod 4$.
So for odd $n\ge 5$, we add $1-0\equiv 1$ to the sum and for even $n$ we add  $0-1\equiv 3$ to the sum, giving us $1007\cdot 1 + 1006 \cdot 3 = 4025$. 
Then in the first four terms, the odd terms are unaffected -- $1-0\cdot k \equiv 1$ no matter what $k$ is -- and the two even terms $n = 2,4$ contribute $1$ and $2$ to the sum respectively (from $k=3,2$ in $0-1\cdot k$) , giving a total of $4030$ for the whole sum.
