Prove that for any integer, $n^2 + 5$ is not divisible by $4$. So I got that there is two cases: odd or even. 
If odd then say $n^2$ is $(2k+1)^2 = 4k^2 + 4k + 1.$ 
then $4k^2 + 4k + 1 + 5$ would need to be divisible by 4 and I don't know where to go from there. 
 A: If $n^2 + 5$ is divisible by $4$, then $3$ is a  square in $\mathbb Z_4$, which is impossible because the only squares in $\mathbb Z_4$ are $0$ and $1$.
A: This may be the easiest way to solve it (this actually uses Ittay's idea, but I go a little bit more into detail). 
Odd: Suppose $n$ is odd; that is, suppose $n=2\ell+1$, where $\ell\in\mathbb{Z}$. Then we have that
$$
n^2+5=(2\ell+1)^2+5=4\ell^2+4\ell+1+5=4(\ell^2+\ell+1)+2,
$$
and this clearly cannot be divisible by $4$.
Even: Suppose $n$ is even; that is, suppose $n=2\ell$, where $\ell\in\mathbb{Z}$. Then we have that
$$
n^2+5=(2\ell)^2+5=4\ell^2+5=2(2\ell^2+2)+1=2m+1,
$$
where $m\in\mathbb{Z}$. Thus, when $n$ is even, we can see that $n^2+5$ is not divisible by $4$.
Hence, when $n$ is either even or odd, the quantity $n^2+5$ will not be divisible by $4$.
A: If $n$ is even, write $n=2k$, and now compute $n^2+5$ in terms of $k$ to see that it is an odd number, thus not a multiple of $4$. If $n$ is odd, then write $n=2k+1$, compute again, and look at what you have, and find out why that can't be a multiple of $4$. 
A: $n$ is even or odd.
$n$ is even: $n = 2m \Rightarrow n^2 + 5 = 4m^2 + 5 = 4(m^2 + 1) + 1$, which isn't divisible by $4$.
$n$ is odd: $n = 2m+1 \Rightarrow n^2 + 5 = 4m^2 + 4m + 1 + 5 = 4(m^2 + m + 1) + 2$, which isn't divisible by $4$. 
A: When $n$ is even,$n^2$ should also be even.Then $n^2+5$ should be odd,so it cannot be divided by 4.
When $n$ is odd,say $n=2k+1$,then $n^2+5=(2k+1)^2+5=4k^2+4k+6$.Since $4k^2$ and $4k$ can be divided by 4,this simply is about whether 6 can be divided by 4. 
A: For a beginning math level, elementary proof, just test case using the fact that every integer has a remainder of $0,1,2$, or $3$ when divided by $4$. (otherwise known as using  congruences)
$n=4k, 4k+1, 4k+2, 4k+3$.
For example, clearly $n$ cannot be a multiple of $4$.
$n=4k$ implies $n^2 +5 = 16k^2 + 4 + 1 = 4(4k^2 +1) + 1$. Remainder 1 ... next case
A: It is also possible to use a reduction ad absurdum argument, which then avoids the use of cases.  Suppose $ \ n^2 \ + \ 5 \ $ were divisible by 4 .  Then $ \ n^2 \ + \ 1 \ $ must be as well.  This requires that $ \ n^2 \ $ and therefore  $ \ n \ $ be odd.  The remainder of the proof is as discussed by almost everyone else here:  $ \ n^2 \ $ must then have the form $ \ (2m \ + \ 1)^2 \ $ [ $ \ m \ $ an integer] , giving $ \ n^2 \ + \ 1 \ = \ (4m^2 \ + \ 4m) \ + \ 2 \ $ , so in fact, this is not divisible by 4 .  Hence, $ \ n^2 \ + \ 5 \ $ isn't either.
You were actually looking at the result that would give you the proof, since $ \ 4 k^2 \ + \ 4k \ + \ 1 \ + \ 5 \ = \ (4 k^2 \ + \ 4k \ + \ 4) \ + \ 2 \ $ likewise has the remainder of 2 .
A: You're halfway there, you just need to take your rewriting to the next logical step. If $n$ is odd, then $n^2 = 4k^2 + 4k + 6$ (all I did there was add the $1$ and the $5$ you had already come up with). Clearly $4k^2 + 4k$ is even and divisible by $4$, and so is $4k^2 + 4k + 4$. But $4k^2 + 4k + 6$ is even but not divisible by $4$.
The even case is much easier: if $n$ is even, then $n^2 + 5$ is odd.
A: $n = 2q\!+\!r,\,\ r=\color{#0a0}0\,$ or $\,\color{#c00}1,\ $ so $\ 4\mid n^2\!+\!5 = 4(q^2\!+\!qr)+r^2\!+\!5\,\Rightarrow\,4\mid \color{#0a0}0^2\!+\!5\,$ or $\,4\mid \color{#c00}1^2\!+\!5\,\Rightarrow\Leftarrow$
