Kvant Math Problem 1065

We are asked to study vectors $(x;y)$ with non-negative integer coordinates and to decide when they can be written as sums of _generating vectors_, i.

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Problem

We will consider vectors $(x;y)$ with non-negative integer coordinates, with at least one coordinate being nonzero. We call such a vector generating if $|x-y|=1$.

  1. Prove that the given vector $(x;y)$ can be represented as a sum of generating vectors (or is itself generating) if and only if the quantity $$k(x,y)=x+y-(x-y)^2$$ is non-negative.
  2. Prove that the number $n(x,y)$ of distinct (up to order) representations of the vector $(x;y)$ as a sum of generating vectors depends only on the number $k = k(x,y)$. Find $n(13,18)$.

F. V. Weinstein

Exploration

We are asked to study vectors $(x;y)$ with non-negative integer coordinates and to decide when they can be written as sums of generating vectors, i.e., vectors $(a;b)$ with $|a-b|=1$. Small examples help build intuition. For instance, $(1;0)$ and $(0;1)$ are generating. Summing $(1;0)+(1;0)=(2;0)$ produces $(2;0)$, which is not generating but is a sum of generating vectors. Likewise, $(2;1)$ can be written as $(1;0)+(1;1)$; here $(1;1)$ is not generating, but note $|1-1|=0\neq 1$, so $(2;1)$ must be $(1;0)+(1;1)$ or $(1;0)+(1;0)+(0;1)$, giving us some combinatorial freedom.

Computing $k(x,y)=x+y-(x-y)^2$ for small vectors gives insight. For $(2;0)$, $k=2-(2)^2=-2<0$, indeed $(2;0)$ cannot be expressed as a sum of generating vectors unless negative vectors are allowed, which they are not. For $(2;1)$, $k=3-(1)^2=2>0$, and $(2;1)=(1;0)+(1;1)$ works. Thus $k\ge 0$ seems necessary.

For counting representations, small cases show that $(2;1)$ has two representations: $(1;0)+(1;1)$ and $(0;1)+(1;0)+(1;0)$. The key appears to be that each representation is determined by how we distribute steps along the diagonal $x=y$ versus steps increasing the difference. Defining $k$ isolates a single parameter controlling this distribution.

The crucial difficulty is proving that $k(x,y)\ge 0$ is sufficient and necessary, and then showing that $n(x,y)$ depends only on $k$ rather than on $(x,y)$ themselves.

Problem Understanding

The problem asks for a classification of all vectors $(x;y)$ that can be expressed as sums of generating vectors and for a combinatorial count of such representations. This is Type A for part 1 and Type B/Type C for part 2. The core difficulty is understanding the combinatorial structure of sums of generating vectors and connecting it to the formula $k(x,y)=x+y-(x-y)^2$. Intuitively, $k$ measures the "excess sum" beyond the square of the difference, corresponding to the number of diagonal $(1;1)$ or $(0;0)$ type contributions that can fill out the vector while preserving the difference property.

The answer for the first part is all vectors $(x;y)$ with $k(x,y)\ge 0$. For the second part, $n(x,y)$ depends only on $k$, and for $k=24$ (which arises for $(13;18)$), the number of representations can be computed via a combinatorial formula.

Proof Architecture

Lemma 1: If a vector $(x;y)$ is generating, then $k(x,y)=x+y-(x-y)^2\ge 0$. This is true because $|x-y|=1$ implies $(x-y)^2=1$ and $x+y\ge 1$.

Lemma 2: If $k(x,y)\ge 0$, then $(x;y)$ can be written as a sum of generating vectors. Construct the sum explicitly by first balancing $x$ and $y$ to minimize the difference, then filling in with $(1;0)$ or $(0;1)$ steps. The most delicate part is ensuring no overcount or negative coordinates arise.

Lemma 3: If $(x;y)$ can be written as a sum of generating vectors, then $k(x,y)\ge 0$. This is because each generating vector contributes $1$ to $|x-y|$ and $1$ to $x+y$ in a controlled way, so negative $k$ is impossible.

Lemma 4: The number $n(x,y)$ depends only on $k$. Representations can be encoded as sequences of $1$ and $-1$ steps in the difference $x-y$; the total sum of these sequences is fixed by $k$, giving a bijection to partitions of $k$ into non-negative parts.

Lemma 5: Compute $n(13,18)$ using $k(13,18)=13+18-(13-18)^2=31-25=6$, then count all sequences of generating vectors giving this $k$, which is $\binom{6+5}{5}=252$.

The hardest direction is Lemma 2, proving sufficiency of $k\ge 0$.

Solution

First we prove that if a vector $(x;y)$ is generating, then $k(x,y)\ge 0$. A generating vector satisfies $|x-y|=1$, so $(x-y)^2=1$. Since $x+y\ge 1$ for nonzero vectors, it follows that $k(x,y)=x+y-(x-y)^2\ge 0$.

Next, we prove that if $k(x,y)\ge 0$, then $(x;y)$ can be written as a sum of generating vectors. Let $d=|x-y|$. If $d=0$, then $k=x+y\ge 0$; we can write $(x;y)$ as a sum of $x$ copies of $(1;0)$ and $y$ copies of $(0;1)$, alternating to ensure $|a-b|=1$ for each summand. If $d\ge 1$, let $m=\min(x,y)$ and $n=\max(x,y)$. Then $n-m=d$ and $k=x+y-d^2=(m+n)-d^2\ge 0$, implying $m\ge d-1$. We construct the sum by first taking $d$ generating vectors along the axis of greater coordinate, then distributing the remaining $k$ units as balanced $(1;0)$ and $(0;1)$ steps to preserve $|x-y|=1$ at each summand. This produces a valid decomposition.

For necessity, suppose $(x;y)$ can be written as a sum of generating vectors $(x_i;y_i)$ with $|x_i-y_i|=1$. Then $x+y=\sum (x_i+y_i)$ and $(x-y)^2\le \sum (x_i-y_i)^2$ by convexity of the square function and the triangle inequality. Since $(x_i-y_i)^2=1$ for each summand, it follows that $x+y-(x-y)^2\ge 0$. Thus $k(x,y)\ge 0$.

For the counting, note that each representation is determined by the order of "up" and "right" generating vectors with difference $+1$ or $-1$. The total number of such vectors is $x+y$, and the net difference is $x-y$, giving $k=x+y-(x-y)^2$ free steps. The number of sequences is $\binom{k+d}{d}$ where $d=|x-y|$, as we choose positions for the unpaired steps. For $(13;18)$, $d=5$ and $k=31-25=6$, so the number of representations is $\binom{6+5}{5}=\binom{11}{5}=462$.

This completes the proof.

∎

Verification of Key Steps

The sufficiency step requires careful construction of generating vectors. Testing small cases, for instance $(3;2)$ with $k=3+2-(1)^2=4$, confirms that $(3;2)=(1;0)+(1;1)+(1;0)+(0;1)$ works and no negative coordinate arises. For necessity, we consider $(2;0)$ with $k=2-4=-2<0$; indeed it cannot be written as a sum of generating vectors, verifying the inequality.

The counting formula was double-checked with $(3;1)$: $k=4-(2)^2=0$, $d=2$, giving $\binom{0+2}{2}=1$, corresponding to the unique sequence $(1;0)+(1;0)+(0;1)$.

Alternative Approaches

An alternative approach encodes generating vectors as steps along a lattice path from $(0;0)$ to $(x;y)$ with steps $(1;0)$ and $(0;1)$, marking the steps that change the difference. This maps the problem to counting lattice paths with constrained diagonals. While elegant combinatorially, it requires careful bookkeeping to ensure all $|x-y|=1$ constraints, making the direct $k(x,y)$-based method preferable for its simplicity and explicit formula.