Integer partitions: number of even parts I've got an elementary, combinatoric question: If the number n is odd, why is the number of even parts = the number of times where each part appears an even number of time = 0 ? 
I mean: Of course, the claim is true (one can show this with any concrete example), but how can you show this in general? 
Thanks for any hint! 
Example: if n = 5
Partitions of 5: 5, 4+1, 3+2, 3+1+1, 2+2+1, 2+1+1+1, 1+1+1+1+1
The number of times where only even parts appears is 0.
The number of times where each part appears an even number of times is also 0.
But: How can I show the claim generally? 
 A: An odd number cannot be partitioned solely into even parts. That's because
a sum of even numbers is even.
Also if a partition has the property that each part occurs an even
number of times, then the total of all parts of the same size is even,
so the total of all parts must be even, and so cannot be odd.
A: We can split this into two parts and generalise.

... why is the number of even parts = the number of times where each part appears an even number of time ...?

Consider an arbitrary integer $k$. The partitions of $n$ into parts which are all multiples of $k$ are in bijection (via transpose of the partition) with the partitions of $n$ into parts which all have frequencies which are multiples of $k$.

... = $0$ ?

The sum of parts which are all multiples of $k$ is also a multiple of $k$, so if $n \not\equiv 0\pmod k$ there are no partitions of $n$ into parts which are all multiples of $k$.
A: Let us consider partitions of $(2n+1)$ into k even parts.
Then,
$(2n+1)=2k_1+2k_2+\cdot 2k_k
          = 2(k_1+k_2+\cdot +k_k)
          =$even number.
Which is impossible-is the first part of the answer .
For the second part,
Let ,
$(2n+1)=2n_1k_1+2n_2k_2+\cdot +2n_kk_k
           =2(n_1k_1+n_2k_2+ \cdot +n_kk_k)
          =$even number.
Which is also impossible .
Hence in both the cases , number of partitions is zero.
