A student asked me the other day the following rather homework-looking question: Given a natural number , how many solutions does the equation

have for and natural numbers?

The question has a very easy answer: Simply notice that and that any like this determines a unique such that is a solution. So, there are solutions if is even (as can be any of ), and there are solutions if is odd.

I didn’t tell the student what the answer is, but I asked what he had tried so far. Among what he showed me there was a piece of paper in which somebody else had scribbled

which caught my interest, and is the reason for this posting.

I don’t think the student had seen the connection between this product of two series, let’s call it , and his question. If we denote by the number of solutions to the equation, the series is the generating function of the sequence , i.e.,

To see this, notice that the coefficient of in is precisely the number of ways we can write as a product of a term from the first series and a term from the second one, i.e., it is the number of solutions with and natural numbers to the equation or, equivalently, . That is to say, the coefficient of in is exactly .

This gives us a purely algebraic (analytic?) way of solving the question, even if there is no understanding of how to approach it from a combinatorial point of view.

Both series on the product that makes up are geometric series, so we have

that of course coincides with the formula we obtained earlier by combinatorial considerations.

The question the student had is a very simple example of a problem about integer partitions, a beautiful area of mathematics that I hope I am not misconstruing by thinking of as a branch of combinatorial number theory. The technique of generating functions is a very useful and powerful combinatorial tool that I have always found quite nice although, granted, its use is a bit of an overkill for the question at hand. At the same time, this technique provides us with a (standard) method for solving any problem of the same kind: For fixed natural numbers , find for each the number of tuples of natural numbers such that

One can then go further to study the much subtler partition function and its relatives.

(And I still don’t know the name of the student, who didn’t bother to introduce himself, and I have no idea who suggested to him to look at to begin with.)

Well, I got the partial solution finally, here is the question that I tried to solve by the equation x + 2y = n, x,y,n > 0

Let n be a positive integer. Harry’s school year has n school days. Harry has budget of exactly $n for buying exactly one snack per day at school. There are only two types of snacks available: M&M for $1. 00 per packet, or a pair of bananas at $2.00 per pair. The following restrictions must apply to Harry.

(1) Harry must spend all $n on snacks during the school year.

(2) Harry does not have to buy a snack each school day.

At the end of the school year, Harry must report how many times he bought M&M, and how many times he bought bananas. How many different reports are possible?

Anyway, thank you for your posting 🙂 It helps me to understand the solution better.

I assume by $\aleph$ you mean $\mathfrak c$, the cardinality of the continuum. You can build $D$ by transfinite recursion: Well-order the continuum in type $\mathfrak c$. At stage $\alpha$ you add a point of $A_\alpha$ to your set, and one to its complement. You can always do this because at each stage fewer than $\mathfrak c$ many points have been selected. […]

Stefan, "low" cardinalities do not change by passing from $L({\mathbb R})$ to $L({\mathbb R})[{\mathcal U}]$, so the answer to the second question is negative. More precisely: Assume determinacy in $L({\mathbb R})$. Then $2^\omega/E_0$ is a successor cardinal to ${\mathfrak c}$ (This doesn't matter, all we need is that it is strictly larger. T […]

The answer is no in general. For instance, by what is essentially an argument of Sierpiński, if $(X,\Sigma,\nu)$ is a $\sigma$-finite continuous measure space, then no non-null subset of $X$ admits a $\nu\times\nu$-measurable well-ordering. The proof is almost verbatim the one here. It is consistent (assuming large cardinals) that there is an extension of Le […]

(As I pointed out in a comment) yes, partial Woodinness is common in arguments in inner model theory. Accordingly, you obtain determinacy results addressing specific pointclasses (typically, well beyond projective). To illustrate this, let me "randomly" highlight two examples: See here for $\Sigma^1_2$-Woodin cardinals and, more generally, the noti […]

I am not sure which statement you heard as the "Ultimate $L$ axiom," but I will assume it is the following version: There is a proper class of Woodin cardinals, and for all sentences $\varphi$ that hold in $V$, there is a universally Baire set $A\subseteq{\mathbb R}$ such that, letting $\theta=\Theta^{L(A,{\mathbb R})}$, we have that $HOD^{L(A,{\ma […]

The question is asking to find all polynomials $f$ for which you can find $a,b\in\mathbb R$ with $a\ne b$ such that the displayed identity holds. The concrete numbers $a,b$ may very well depend on $f$. A priori, it may be that for some $f$ there is only one pair for which the identity holds, it may be that for some $f$ there are many such pairs, and it may a […]

The reflection principle is a theorem schema in ZFC, meaning that for each formula $\phi(\vec x)$ we can prove in ZFC a version of the principle for $\phi$. In particular, it gives us that if $\phi$ holds (in the universe of sets) then there is some ordinal $\alpha$ such that $V_\alpha\models \phi$. It follows from this that (assuming its consistency) $\math […]

All proofs of the Bernstein-Cantor-Schroeder theorem that I know either directly or with very little work produce an explicit bijection from any given pair of injections. There is an obvious injection from $[0,1]$ to $C[0,1]$ mapping each $t$ to the function constantly equal to $t$, so the question reduces to finding an explicit injection from $C[0,1]$ to $[ […]

One way we formalize this "limitation" idea is via interpretative power. John Steel describes this approach carefully in several places, so you may want to read what he says, in particular at Solomon Feferman, Harvey M. Friedman, Penelope Maddy, and John R. Steel. Does mathematics need new axioms?, The Bulletin of Symbolic Logic, 6 (4), (2000), 401 […]

"There are" examples of discontinuous homomorphisms between Banach algebras. However, the quotes are there because the question is independent of the usual axioms of set theory. I quote from the introduction to W. Hugh Woodin, "A discontinuous homomorphism from $C(X)$ without CH", J. London Math. Soc. (2) 48 (1993), no. 2, 299-315, MR1231 […]

I’m the student asked the question 😉

Well, I got the partial solution finally, here is the question that I tried to solve by the equation x + 2y = n, x,y,n > 0

Let n be a positive integer. Harry’s school year has n school days. Harry has budget of exactly $n for buying exactly one snack per day at school. There are only two types of snacks available: M&M for $1. 00 per packet, or a pair of bananas at $2.00 per pair. The following restrictions must apply to Harry.

(1) Harry must spend all $n on snacks during the school year.

(2) Harry does not have to buy a snack each school day.

At the end of the school year, Harry must report how many times he bought M&M, and how many times he bought bananas. How many different reports are possible?

Anyway, thank you for your posting 🙂 It helps me to understand the solution better.

Hi Eliot,

I’m glad this helped. Marion mentioned to me the `M&Ms problem’ the other day, I figured this was the same question without distractions.