Kvant Math Problem 642
The coefficients are restricted to the set ${-1,0,1}$, and two neighboring coefficients cannot both be nonzero.
Verified: no
Verdicts: SKIP + SKIP
Solve time: 5m12s
Source on kvant.digital
Problem
Prove that every positive integer can be represented in the form $a_0+2a_1+2^2a_2+\ldots+2^na_n$, where each of the numbers $a_k=0$, $-1$ is either 1 and $a_k\cdot a_{k+1}=0$ for all $0\le k\le n-1$, and moreover this representation is unique.
I. K. Zhuk
Exploration
The coefficients are restricted to the set ${-1,0,1}$, and two neighboring coefficients cannot both be nonzero. This resembles a balanced binary representation with an additional nonadjacency condition.
It is useful to compute the first few examples:
$$1=1,$$
$$2=2,$$
$$3=4-1,$$
$$4=4,$$
$$5=4+1,$$
$$6=8-2,$$
$$7=8-1,$$
$$8=8.$$
The pattern suggests a recursive rule. If $N$ is even, then $N=2M$. A representation of $M$ immediately yields a representation of $N$ by shifting all coefficients one place to the left.
For odd $N$, there are two possibilities modulo $4$.
If $N\equiv1\pmod4$, then $N-1$ is divisible by $4$, so
$$N=1+4K.$$
If $N\equiv3\pmod4$, then $N+1$ is divisible by $4$, so
$$N=-1+4K.$$
In either case the constant coefficient is determined uniquely by the residue modulo $4$, and the remaining part is divisible by $4$. This is promising because divisibility by $4$ forces the coefficient of $2$ to be $0$, automatically satisfying the nonadjacency condition near $a_0$.
The crucial point is uniqueness. A naive parity argument determines $a_0$, but one must then show that after removing $a_0$ and dividing by $4$, the remaining coefficients again satisfy the same admissibility condition. If this reduction is not checked carefully, the induction breaks.
Problem Understanding
We must prove that every positive integer admits a representation
$$N=a_0+2a_1+2^2a_2+\cdots+2^na_n,$$
where every coefficient belongs to ${-1,0,1}$ and
$$a_k a_{k+1}=0 \qquad (0\le k<n),$$
and that this representation is unique.
This is a Type B problem. The statement to be proved consists of both existence and uniqueness.
The core difficulty is finding a recursive structure that preserves the nonadjacency condition. Congruences modulo $4$ provide exactly such a structure, because the residue class of $N$ modulo $4$ determines the first coefficient and forces the next coefficient to be $0$.
Proof Architecture
First, prove existence by strong induction on $N$.
For even $N$, write $N=2M$ and shift a representation of $M$ one place to the left.
For odd $N$, determine $a_0$ from the residue of $N$ modulo $4$, write $N=a_0+4K$, represent $K$ inductively, and place a zero coefficient at the power $2$.
Next, prove uniqueness by induction on $N$.
Show that in any admissible representation of an odd number, the coefficient $a_0$ is uniquely determined by the residue modulo $4$: it equals $1$ when $N\equiv1\pmod4$ and equals $-1$ when $N\equiv3\pmod4$.
After removing $a_0$ and dividing by $4$, obtain a smaller integer whose coefficients again form an admissible representation.
For even numbers, parity forces $a_0=0$; then dividing by $2$ reduces uniqueness to a smaller integer.
The most delicate lemma is that, after determining $a_0$ for an odd number, the coefficient $a_1$ must be $0$, so division by $4$ produces an admissible representation of a smaller integer.
Solution
We prove existence and uniqueness simultaneously by induction on the positive integer $N$.
For $N=1$ we have
$$1=1,$$
which satisfies all requirements. The representation is unique because the only admissible sum of absolute value at most $1$ equal to $1$ is the single coefficient $a_0=1$.
Assume that every positive integer less than $N$ has a unique admissible representation. We prove the same for $N$.
First we establish existence.
Suppose that $N$ is even. Then $N=2M$ with $M<N$. By the induction hypothesis,
$$M=b_0+2b_1+\cdots+2^rb_r,$$
where each $b_i\in{-1,0,1}$ and $b_i b_{i+1}=0$.
Multiplying by $2$ gives
$$N=2b_0+2^2b_1+\cdots+2^{r+1}b_r.$$
Thus $N$ has the representation
$$a_0=0,\qquad a_{i+1}=b_i.$$
The coefficients still belong to ${-1,0,1}$, and
$$a_i a_{i+1}=b_{i-1}b_i=0$$
for all relevant indices. Hence an admissible representation exists.
Now suppose that $N$ is odd.
If $N\equiv1\pmod4$, then
$$N=1+4K$$
for some integer $K\ge0$.
If $K=0$, then $N=1$, already treated.
If $K>0$, then $K<N$. By the induction hypothesis,
$$K=b_0+2b_1+\cdots+2^rb_r.$$
Therefore
$$N =1+4(b_0+2b_1+\cdots+2^rb_r) =1+2^2b_0+2^3b_1+\cdots+2^{r+2}b_r.$$
This corresponds to
$$a_0=1,\qquad a_1=0,\qquad a_{i+2}=b_i.$$
The nonadjacency condition holds because $a_1=0$ and the sequence $(b_i)$ already satisfies it.
If $N\equiv3\pmod4$, then
$$N=-1+4K$$
with $K=(N+1)/4<N$.
Using an admissible representation of $K$ and arguing exactly as above, we obtain
$$N=-1+2^2b_0+2^3b_1+\cdots+2^{r+2}b_r,$$
which is admissible.
Existence is proved.
We now prove uniqueness.
Suppose first that $N$ is even and
$$N=a_0+2a_1+2^2a_2+\cdots$$
is an admissible representation.
Since all terms except $a_0$ are even and $N$ is even, the coefficient $a_0$ must be $0$.
Hence
$$N=2(a_1+2a_2+2^2a_3+\cdots).$$
Dividing by $2$ yields
$$\frac N2=a_1+2a_2+2^2a_3+\cdots .$$
The coefficients on the right are admissible. Since $N/2<N$, the induction hypothesis gives a unique representation of $N/2$. Consequently the representation of $N$ is unique.
Now suppose that $N$ is odd and
$$N=a_0+2a_1+2^2a_2+\cdots$$
is admissible.
Because $N$ is odd, the coefficient $a_0$ is odd. Since $a_0\in{-1,0,1}$,
$$a_0=\pm1.$$
The condition $a_0a_1=0$ then forces
$$a_1=0.$$
Hence
$$N=a_0+4(a_2+2a_3+2^2a_4+\cdots).$$
Reducing modulo $4$ gives
$$N\equiv a_0\pmod4.$$
Therefore
$$a_0= \begin{cases} 1,&N\equiv1\pmod4,\ -1,&N\equiv3\pmod4. \end{cases}$$
Thus $a_0$ is uniquely determined by $N$.
After subtracting $a_0$ and dividing by $4$ we obtain
$$\frac{N-a_0}{4} = a_2+2a_3+2^2a_4+\cdots .$$
The coefficients on the right still satisfy the admissibility condition. Since
$$\frac{N-a_0}{4}<N,$$
the induction hypothesis yields a unique representation of this smaller integer. Hence all coefficients $a_2,a_3,\ldots$ are uniquely determined, and so the representation of $N$ is unique.
The induction is complete. Every positive integer possesses exactly one admissible representation of the required form.
This completes the proof.
∎
Verification of Key Steps
The first delicate step is the determination of $a_0$ for odd numbers. Since $a_0=\pm1$, the condition $a_0a_1=0$ forces $a_1=0$. Then
$$N=a_0+4T,$$
so $N\equiv a_0\pmod4$. Without proving $a_1=0$, one would only obtain
$$N\equiv a_0+2a_1\pmod4,$$
and $a_0$ would no longer be determined uniquely.
The second delicate step is the reduction after dividing by $4$. From
$$N=a_0+2^2a_2+2^3a_3+\cdots,$$
we obtain
$$\frac{N-a_0}{4} =a_2+2a_3+\cdots.$$
The new coefficient sequence is $(a_2,a_3,\ldots)$. Since the original sequence satisfied
$$a_k a_{k+1}=0$$
for every $k$, the shifted sequence satisfies the same condition. This must be checked explicitly before applying the induction hypothesis.
The third delicate step is the even case. Parity alone gives $a_0=0$. After division by $2$ the remaining coefficients are $(a_1,a_2,\ldots)$, and the admissibility condition survives unchanged. Forgetting to verify this would leave the induction unjustified.
Alternative Approaches
A different proof starts from ordinary binary expansion. Whenever two consecutive binary digits equal $1$, replace the block
$$2^k+2^{k+1}$$
by
$$-2^k+2^{k+2}.$$
Repeatedly performing this transformation removes all adjacent nonzero digits. One must then prove that the process terminates and that the resulting representation is independent of the order of transformations. Uniqueness can be established by showing that two distinct reduced representations would yield a nontrivial admissible representation of zero, which is impossible.
The inductive modulo $4$ approach is preferable because existence and uniqueness arise from the same recursive decomposition. The residue class modulo $4$ immediately identifies the lowest nonzero coefficient and reduces the problem to a strictly smaller integer.