Partition of odd $n$ into summands

Question

For odd $n$, explain why there can be no partition of $n$ into summands such that each part appears an even number of times.

Please help me, I don't have any idea

Answer 1

Imagine we have a possibly large number $n$ of candies, We also have a group of children, number unspecified. We give out (all) the candies to the children, perhaps in a very unfair way. This partitions the candies, it expresses the number of candies as a sum.

Suppose that for any $k$, the number of children who get $k$ candies is even. (Recall that in particular $0$ is even. So if $0$ of the kids get exactly $7$ candies, that's consistent with our condition.)

We will show that the total number of candies must be even.

Consider for example the $1$-candy kids. Since the number of such kids is even, between them they must have an even number of candies.

Now consider the $2$-candy kids. It actually doesn't matter whether there is an even number of such kids. In any case, the number of candies among these kids is even.

Now consider the $3$-candy kids. There is an even number of them, so between them they must have an even number of candies. For instance, if there are $14$ $3$-candy kids, they account for the even number $42$ candies.

And so on.

So the $1$ candy kids account for an even number of candies. So do the $2$-candy kids, and the $3$-candy kids, and so on. The total number of candies is obtained by adding up the candies of the $1$-candy kids, the $2$-candy kids, and so on. Since the sum of a bunch of even numbers is even, the total number of candies must be even. Thus $n$ cannot be odd.

Remark: Pictures would help a lot. Instead look at the sample partition $$(7+7) + (6+6+6+6) +(5+5+5+5+5+5)+(3+3+3+3+3+3+3+3)+(1+1+1+1).$$ We have used parentheses to make things clear. The sum of the first group is even, as is the sum of the second group, and the third, and so on, so the total must be even.

One can also give a quick proof by formula. Suppose the part $1$ appears $2k_1$ times, and the part $2$ appears $2k_2$ times, and so on until the part $m$ which appears $2k_m$ times. Then $$n=(1)(2k_1)+(2)(2k_2)+(3)(2k_3)+\cdots +(m)(2k_m)=2\left(k_1+2k_2+3k_3+\cdots +mk_m\right),$$ so $n$ is even.