Show $X=\{n \in \mathbb{N}: \text{n is odd and} \ n = k(k+1) \text{for some} \ k \in \mathbb{N}\}=\emptyset$ Show $X=\{n \in \mathbb{N}: \text{n is odd and} \ n = k(k+1) \text{for some} \ k \in \mathbb{N}\}=\emptyset$
My proof is as follow, please point if I have made any mistake. 
proof:
we have $\emptyset \subseteq X$
suppose $X≠\emptyset$  pick $n \in X$
Then there are 2 cases
1st case: n is odd then n=(k+1)k
Then 
suppose k is odd $\implies$ k+1 is even $\implies$ n is even
2nd case: consider k is even $\implies k+1$ is odd
then 
n=(k+1)k for some $k \in \mathbb{N}=\emptyset \implies n$ is even
Therefore, n is neither even nor odd, so $k \in \mathbb{N} \implies n \not\in X$
and $\implies X= \emptyset$     
Q.E.D
 A: Your proof is correct, I would just be a bit pedantic about your format and ordering. So I'll follow on from your proof:
Suppose for contradiction that $X\neq \emptyset$ and pick $n\in X$
Then we know $n$ is odd and we can write $n=k(k+1)$ for some $k\in\mathbb{N}$
Then the two cases are:
$(i)\quad k\equiv 0\bmod 2\implies n\equiv 0\bmod 2\implies n\not\in X$
$(ii)\,\,\,\,k\equiv 1\bmod 2\implies k+1\equiv 0\bmod 2\implies n\equiv 0\bmod 2\implies n\not\in X$
Therefore by exhaustion of the all the possible cases, we have arrived at a contradiction, and hence $X=\emptyset$. 
It's exactly the same as what you did, but just a bit neater (in my opinion).
A: An even shorter proof (not that yours is wrong): 
One of $k, k+1$ must be even, therefore so is the product $k(k+1)$.
A: Here is a combinatorial proof that $k(k+1)$ is always even for $k\in\mathbb{N}$. The number of two element subsets of a $k+1$ element set is 
$$
\binom{k+1}{2}=\frac{k(k+1)}{2}
$$
as there are $k+1$ choices for the first element of the subset and $k$ choices for the second element of the subset, but as $\{a,b\}$ and $\{b,a\}$ represent the same set, we divide the product by $2$ to find the number of such subsets. The result follows.
A: You are taking a very long-winded approach.
Your $X$ can be expressed as:
$$X=\{n\in\mathbb N|\text{n is odd} \land \text{n is even}\}$$
and we need to prove $X=\emptyset$.
