Prove that if $n$ is an integer, then $n^2 + n^3$ is an even number I am trying to work through some of the problems in Stephen Lay's Introduction to Analysis with Proof before my Real Analysis class in the fall term starts, and I was just wondering if I could get some feedback as to whether or not I have completed this proof properly
The question:

Prove that if $n$ is an integer, then $n^2 + n^3$ is an even number

 A: Your approach for the proof is incorrect. You want to prove an implication $P \to Q$. So if you assume that $Q$ is true, then you have nothing left to prove. You should just begin with an integer $n$ then try to see what can be said about $n^3+n^2$. For example, you can say
$$n^3+n^2=n^2(n+1).$$
Now if $n$ is odd then $n+1$ is even and vice versa. So you have a product of odd number and even number and hence even. 
A: Hint
$$n^3+n^2=n(n(n+1))$$
$n(n+1)$ is the product of two consecutive integers.
A: Alternatively, by Fermat's Little Theorem,
$$\begin{align}n^2 + n^3 &= n^2 + n\cdot n^2\\
&\equiv n + n\cdot n \pmod  2\\
&= n + n^2 \pmod 2\\
&\equiv n + n \pmod 2\\
&= 2n \pmod 2\\
&\equiv 0 \pmod 2\end{align}$$
A: Here's a shorter proof:  Select any $n \in \mathbb{Z}$.  There are two cases:  $n \equiv 0, 1 \pmod{2}$.
In either case, use modular arithmetic to compute $n^2 + n^3 \pmod{2}$.  What can we conclude?
A: My attempt via unique prime factorization of integers:
$n = p_1^{a_1} p_2^{a_2} \dots p_n^{a_n}$ for some unique choice of (positive) prime $p_1, \dots, p_n$ and natural $a_1, \dots, a_n$.
Case 1:
Let $n$ be odd. Then each $p_i$, with $1 \leq i \leq n$ is odd (i.e., not 2). Now, $n^2 = (p_1^{a_1} p_2^{a_2} \dots p_n^{a_n})^2 = p_1^{2a_1} p_2^{2a_2} \dots p_n^{2a_n}$, which is still odd. Similar for $n^3$. Thus $n^2 + n^3$ is even, being the sum of two odd numbers.
Case 2:
Let $n$ be even. Then we know that one of the $p_i$'s is $2$. We can write our unique factorization of $n$ as $2^{a_1} p_2^{a_2} \dots p_n^{a_n}$. Now it should be obvious that $n^2 = 2^{2a_1} p_2^{2a_2} \dots p_n^{2a_n}$ is even, since it's a multiple of 2, and likewise for $n^3$. Thus we see that $n^2 + n^3$ is the sum of two even integers and is therefore even.
A: This came up as a question that might have the answer to my question. It doesn't, but since I'm here, I might as well:
Suppose $n$ is odd. Then $n^2$ is also odd, as well as $n^3$. But two odd numbers add up to an even number, hence $n^2 + n^3$ is even.
If instead $n$ is even, then $n^2$ and $n^3$ are both even also, and since even numbers add up to even numbers anyway, we're done.
A: Integer can be separated into two types one is even number and the other is odd numbers
Case 1: The integer is even:
Even numbers can be represented as: $n = 2k$
$n^2 + n^3 = (2k)^2 + (2k)^3 = 4k^2 + 8k^3 = 2(2k^2 + 4k^3) = 2m$
Therefore if n is even $->$ ${n^2 + n^3}$ is even 
Case 2: The integer is odd:
Odd numbers can be represented as $n = 2k + 1$
$n^2 + n^3 = (2k + 1)^2 + (2k + 1)^3 = 8k^3 + 16k^2 + 10k + 2 = 2(4k^3 + 8k^2 + 5k + 1) = 2x$ 
Therefore if $n$ is odd $->$ $n^2 + n^3$ is even 
