CF 213B - Numbers

CF 213B - Numbers

Rating: 1900
Tags: combinatorics, dp
Solve time: 54s
Verified: no

Solution

Problem-Type Check

This is a Type A, “Find all X” problem. A complete solution must do two things:

First, it must show that every listed family of solutions actually satisfies the system.

Second, it must prove that no other solutions exist.

The proposed proof does both parts.

It explicitly constructs eigenvectors corresponding to the admissible values of $y$, and explains the resulting solution spaces. It also proves exhaustiveness by decomposing an arbitrary vector into the eigenbasis and showing that only eigenspace components with eigenvalue $y$ may survive.

So structurally, the proof addresses both directions required for a complete “find all solutions” problem.

Step-by-Step Verification

Step 1: Rewriting the system as $AX=yX$ for a circulant operator $A$ - VALID

The operator

$$A(x_1,\dots,x_5) = (x_2+x_5,\ x_1+x_3,\ x_2+x_4,\ x_3+x_5,\ x_4+x_1)$$

matches the given equations exactly. The reformulation is correct.

Step 2: Claim that the recurrence

$$x_{i+2}=yx_{i+1}-x_i$$

follows from the system - VALID

Each original equation is precisely of this form after rearrangement. For example:

$$x_1+x_3=yx_2 \quad\Rightarrow\quad x_3=yx_2-x_1.$$

Checking cyclic indices explicitly:

For $i=1$:

$$x_3=yx_2-x_1.$$

For $i=2$:

$$x_4=yx_3-x_2.$$

For $i=3$:

$$x_5=yx_4-x_3.$$

For $i=4$:

$$x_1=yx_5-x_4.$$

For $i=5$:

$$x_2=yx_1-x_5.$$

All are consistent.

Step 3: Definition of

$$v_k=(1,\omega^k,\omega^{2k},\omega^{3k},\omega^{4k})$$

with $\omega=e^{2\pi i/5}$ - VALID

This is standard and well-defined.

Step 4: Computation

$$A(v_k)=(\omega^k+\omega^{-k})v_k$$

• VALID

Let the $j$-th coordinate of $v_k$ be $\omega^{(j-1)k}$.

Then the $j$-th coordinate of $A(v_k)$ is

$$\omega^{(j-2)k}+\omega^{jk} = (\omega^{-k}+\omega^k)\omega^{(j-1)k}.$$

Hence every coordinate is multiplied by the same scalar

$$\omega^k+\omega^{-k}.$$

The eigenvalue computation is correct.

Step 5: Identification

$$\lambda_k=\omega^k+\omega^{-k}=2\cos\frac{2\pi k}{5}$$

• VALID

This follows from Euler’s formula:

$$e^{i\theta}+e^{-i\theta}=2\cos\theta.$$

Step 6: Claim that $v_0,\dots,v_4$ form a basis because the Vandermonde determinant is nonzero - VALID

The matrix with columns $v_k$ is

$$(\omega^{ij})_{0\le i,j\le4},$$

whose determinant is

$$\prod_{0\le i<j\le4}(\omega^j-\omega^i).$$

The fifth roots of unity are distinct, so every factor is nonzero. Hence the determinant is nonzero and the vectors are linearly independent.

Since there are five vectors in $\mathbb C^5$, they form a basis.

Step 7: Decomposition

$$X=\sum_{k=0}^4 c_k v_k$$

and application of $A$ - VALID

This follows directly from the basis property and linearity.

Step 8: Deduction that

$$c_k(\lambda_k-y)=0$$

for all $k$ - VALID

From

$$\sum_{k=0}^4 c_k(\lambda_k-y)v_k=0,$$

linear independence implies each coefficient vanishes.

No hidden assumption is used here.

Step 9: Conclusion that all nonzero components lie in eigenspaces with eigenvalue $y$ - VALID

This is exactly what Step 8 says.

Step 10: Computation of the three distinct eigenvalues

$$2,\quad \frac{\sqrt5-1}{2},\quad -\frac{\sqrt5+1}{2}$$

• VALID

Indeed:

$$2\cos\frac{2\pi}{5}=\frac{\sqrt5-1}{2},$$

and

$$2\cos\frac{4\pi}{5}=-\frac{\sqrt5+1}{2}.$$

Step 11: Case $y=2$, obtaining

$$x_1=x_2=x_3=x_4=x_5$$

• VALID

Only the eigenspace of $v_0=(1,1,1,1,1)$ survives.

Substitution check:

$$x_5+x_2=c+c=2c=yx_1=2c.$$

The same holds cyclically.

Step 12: Description of the eigenspace for

$$y=\frac{\sqrt5-1}{2}$$

as

$$a\Re(v_1)+b\Im(v_1)$$

• VALID

Since $\lambda_1=\lambda_4$, the eigenspace over $\mathbb C$ is spanned by $v_1,v_4$, where $v_4=\overline{v_1}$.

The real eigenspace is exactly the span of the real and imaginary parts.

This is complete.

Step 13: Description of the eigenspace for

$$y=-\frac{\sqrt5+1}{2}$$

as

$$a\Re(v_2)+b\Im(v_2)$$

• VALID

Same reasoning as Step 12.

Step 14: Claim that for all other $y$, only the zero solution exists - VALID

If $y$ is not an eigenvalue, then all $c_k=0$. Hence $X=0$.

This includes the possibility $X=0$, which indeed satisfies the system for every $y$.

Step 15: Final exhaustion argument using the spectrum - VALID

The proof has already classified every solution through the basis decomposition. The spectral remark is supplementary and correct.

Completeness Check

The proof is complete.

The only subtle point is that the original IMO problem may have intended real variables rather than complex ones. The proof handles this correctly by working over $\mathbb C$ first and then explicitly extracting the real eigenspaces in Cases 2 and 3.

No cases are omitted.

The proof also correctly handles the zero solution. For every value of $y$, the zero vector satisfies the system, and the argument for non-eigenvalue $y$ concludes exactly that the zero solution is the only one.

The transition from complex eigenvectors to real solution spaces is justified correctly through conjugate pairs.

No implicit assumptions remain unproved.

Summary

The argument is logically coherent from beginning to end.

The eigenvector computation is correct, the basis argument is rigorous, and the exhaustion step completely classifies all solutions.

The proof successfully establishes both existence and uniqueness of the listed solution families.

VERDICT: PASS - the proof is complete and all steps are correctly justified.