How to put 9 pigs into 4 pens so that there are an odd number of pigs in each pen?

Since this has been exposed, some claim, to be a riddle and not a bona fide math question, why not completely drive the stake through its heart:

Pen 1: 7 pigs

Pen 2: 1 pig

Pen 3: 3 pigs

Pen 4: -2 pigs

Now the astute reader will note that -2 pigs is a pretty darn odd number of pigs to be in a pen!

It appears to be a trick question.

Make 3 pens, put 3 pigs in each pen. Then put a 4th pen around all 3 of the other ones, and you have 9 pigs in that pen.

Update

@MJD found a source for this problem with solution, see Boys' Life magazine from 1916.

Lemma: $-\frac{1}{12}$ is an integer.

proof: Consider the Riemann zeta function $\zeta$ evaluated at $-1$. By analytic continuation, $\zeta(-1) = -\frac{1}{12}$. However, we also have the series expansion $\zeta(s)=\displaystyle \sum_{n=1}^\infty n^{-s}$, so (ignoring issues with convergence), $\zeta(-1)=1+2+3+\cdots$. This is an infinite sum of integers. Any finite sum of integers is an integer, and the since the integers are a closed set (and hence contain all limit points) this also holds in the limiting case. Hence $-\frac{1}{12}$ is an integer.

Corollary $\frac{2}{3}$ is odd.

proof: By above, since $-\frac{1}{12}$ is an integer, $4 (-\frac{1}{12})=-\frac{1}{3}$ is an even integer. Since the successor of any even integer is odd, $-\frac13 +1 = \frac23$ is odd.

Theorem It is possible to put 9 pigs in 4 pens such that each pen has an odd number of pigs.

proof: For the first pen, put $7$ of the pigs in. Cut the remaining $2$ pigs into equal thirds and put two of the thirds in each of the remaining pens. Since $7=2\times3+1$, $7$ is odd. By corollary above, $\frac23$ is odd. Hence all four pens have an odd number of pigs.

Note: The above is humor.