IMO 1974 Problem 3
Let
Proposed by: -
Verified: no
Verdicts: PASS + FAIL
Solve time: 11m12s
Problem
Prove that the number $\textstyle \sum_{k = 0}^n \tbinom{2n+1}{2k+1} 2^{3k}$ is not divisible by 5 for any integer $n \geq 0$.
Exploration
Let
$$S_n=\sum_{k=0}^{n}\binom{2n+1}{2k+1}2^{3k}.$$
The first five values are
$$S_0=1,$$
$$S_1=\binom31+\binom33\cdot 8=3+8=11,$$
$$S_2=\binom51+\binom53\cdot 8+\binom55\cdot 64 =5+80+64=149,$$
$$S_3=\binom71+\binom73\cdot 8+\binom75\cdot 64+\binom77\cdot 512 =7+280+1344+512=2143,$$
$$S_4=9+84\cdot 8+126\cdot 64+36\cdot 512+4096=30857.$$
Reducing modulo $5$,
$$1,\ 11,\ 149,\ 2143,\ 30857 \equiv 1,1,4,3,2 \pmod 5.$$
None is divisible by $5$.
The sum involves only odd binomial coefficients. Such sums are often extracted from a binomial expansion by taking
$$(a+b)^{2n+1}-(a-b)^{2n+1},$$
since the even powers cancel and the odd powers remain. Choosing $b=\sqrt 8$ would produce the factor $8^k$.
Trying this,
$$(1+\sqrt8)^{2n+1}-(1-\sqrt8)^{2n+1} =2\sqrt8\sum_{k=0}^{n}\binom{2n+1}{2k+1}8^k.$$
Hence
$$S_n=\frac{(1+\sqrt8)^{2n+1}-(1-\sqrt8)^{2n+1}}{2\sqrt8}.$$
The expression suggests introducing a recurrence. Let
$$a_m=\frac{(1+\sqrt8)^m-(1-\sqrt8)^m}{2\sqrt8}.$$
Then $S_n=a_{2n+1}$. Since $1\pm\sqrt8$ are roots of
$$x^2-2x-7=0,$$
the sequence $a_m$ satisfies
$$a_{m+2}=2a_{m+1}+7a_m.$$
Modulo $5$,
$$a_{m+2}\equiv 2a_{m+1}+2a_m.$$
Computing a few terms modulo $5$,
$$a_0=0,\quad a_1=1,$$
$$a_2=2,\quad a_3=1,\quad a_4=1,\quad a_5=4,$$
$$a_6=0,\quad a_7=3,\quad a_8=1,\quad a_9=3,\quad a_{10}=3,\quad a_{11}=2.$$
The pattern repeats with period $12$. The odd-indexed terms in one period are
$$1,1,4,3,3,2,$$
none of which is $0$. This appears to be the key idea.
The step most likely to conceal an error is proving the periodicity modulo $5$. It must be checked rigorously rather than inferred from a few terms.
Problem Understanding
We are given the sequence
$$S_n=\sum_{k=0}^{n}\binom{2n+1}{2k+1}2^{3k},$$
and must prove that $5\nmid S_n$ for every integer $n\ge 0$.
This is a Type B problem. The task is to prove a universal statement.
The objects involved are binomial coefficients and powers of $8$. The restriction to odd lower indices suggests extracting the sum from a suitable binomial expansion. A direct study of the sum modulo $5$ is difficult because many terms are present and the number of terms grows with $n$. The main difficulty is to convert the sum into an expression whose residue modulo $5$ can be analyzed uniformly for all $n$.
The natural route is to represent the sum as a term of a linear recurrence and then study that recurrence modulo $5$.
Proof Architecture
We shall prove the result through the following claims.
Lemma 1
Define
$$a_m=\frac{(1+\sqrt8)^m-(1-\sqrt8)^m}{2\sqrt8}.$$
Then
$$S_n=a_{2n+1}.$$
This follows by expanding both powers with the binomial theorem and observing that the even powers cancel.
Lemma 2
The sequence $(a_m)$ satisfies
$$a_{m+2}=2a_{m+1}+7a_m$$
for all $m\ge0$, with initial values
$$a_0=0,\qquad a_1=1.$$
This follows because $1\pm\sqrt8$ are roots of $x^2-2x-7=0$.
Lemma 3
Modulo $5$, the sequence $(a_m)$ has period $12$.
This is proved by computing twelve consecutive terms from the recurrence and showing that the pair $(a_{12},a_{13})$ coincides with $(a_0,a_1)$ modulo $5$.
Lemma 4
No odd-indexed term of one period is congruent to $0\pmod 5$.
This is verified directly from the computed period.
The most delicate step is Lemma 3, because periodicity must be established rigorously from the recurrence and not merely guessed from numerical data.
Solution
Lemma 1
Let
$$a_m=\frac{(1+\sqrt8)^m-(1-\sqrt8)^m}{2\sqrt8}.$$
Then
$$S_n=a_{2n+1}.$$
Proof
Applying the binomial theorem,
$$(1+\sqrt8)^{2n+1} = \sum_{j=0}^{2n+1} \binom{2n+1}{j}(\sqrt8)^j,$$
and
$$(1-\sqrt8)^{2n+1} = \sum_{j=0}^{2n+1} \binom{2n+1}{j}(-\sqrt8)^j.$$
Subtracting,
$$(1+\sqrt8)^{2n+1}-(1-\sqrt8)^{2n+1} = 2\sum_{k=0}^{n} \binom{2n+1}{2k+1} (\sqrt8)^{,2k+1}.$$
Since
$$(\sqrt8)^{2k+1} = \sqrt8,8^k,$$
we obtain
$$(1+\sqrt8)^{2n+1}-(1-\sqrt8)^{2n+1} = 2\sqrt8 \sum_{k=0}^{n} \binom{2n+1}{2k+1}8^k.$$
Dividing by $2\sqrt8$,
$$a_{2n+1} = \sum_{k=0}^{n} \binom{2n+1}{2k+1}8^k = S_n.$$
Thus $S_n=a_{2n+1}$. ∎
Certification: this identifies the given sum as a specific term of a recurrence sequence; merely observing a resemblance to a binomial expansion would not provide a usable algebraic identity.
Lemma 2
The sequence $(a_m)$ satisfies
$$a_{m+2}=2a_{m+1}+7a_m$$
for all $m\ge0$, and
$$a_0=0,\qquad a_1=1.$$
Proof
Set
$$r_1=1+\sqrt8,\qquad r_2=1-\sqrt8.$$
Both satisfy
$$r_i^2-2r_i-7=0,$$
hence
$$r_i^{m+2}=2r_i^{m+1}+7r_i^m$$
for every $m\ge0$.
Subtracting the corresponding identities for $r_1$ and $r_2$,
$$r_1^{m+2}-r_2^{m+2} = 2(r_1^{m+1}-r_2^{m+1}) + 7(r_1^m-r_2^m).$$
Dividing by $2\sqrt8$,
$$a_{m+2}=2a_{m+1}+7a_m.$$
Also,
$$a_0=\frac{1-1}{2\sqrt8}=0,$$
and
$$a_1 = \frac{(1+\sqrt8)-(1-\sqrt8)}{2\sqrt8} = 1.$$
The lemma follows. ∎
Certification: this produces a recurrence with integer coefficients; working directly with the radical expression would make the modular analysis cumbersome.
Lemma 3
Modulo $5$, the sequence $(a_m)$ has period $12$.
Proof
Reducing the recurrence modulo $5$,
$$a_{m+2}\equiv 2a_{m+1}+2a_m \pmod 5.$$
Starting from
$$a_0\equiv0,\qquad a_1\equiv1,$$
we compute successively:
$$a_2\equiv2,$$
$$a_3\equiv2\cdot2+2\cdot1\equiv1,$$
$$a_4\equiv2\cdot1+2\cdot2\equiv1,$$
$$a_5\equiv2\cdot1+2\cdot1\equiv4,$$
$$a_6\equiv2\cdot4+2\cdot1\equiv0,$$
$$a_7\equiv2\cdot0+2\cdot4\equiv3,$$
$$a_8\equiv2\cdot3+2\cdot0\equiv1,$$
$$a_9\equiv2\cdot1+2\cdot3\equiv3,$$
$$a_{10}\equiv2\cdot3+2\cdot1\equiv3,$$
$$a_{11}\equiv2\cdot3+2\cdot3\equiv2,$$
$$a_{12}\equiv2\cdot2+2\cdot3\equiv0,$$
$$a_{13}\equiv2\cdot0+2\cdot2\equiv1.$$
Hence
$$(a_{12},a_{13})\equiv(0,1)\equiv(a_0,a_1)\pmod5.$$
Since every later term is determined uniquely from the preceding two terms by the recurrence, the sequence from index $12$ onward repeats exactly the sequence from index $0$ onward modulo $5$. Thus the sequence is periodic modulo $5$ with period $12$. ∎
Certification: this establishes periodicity by matching the complete state $(a_m,a_{m+1})$, not merely one term, which would be insufficient.
Lemma 4
No odd-indexed term of the period is congruent to $0\pmod5$.
Proof
From the values computed in Lemma 3,
$$a_1\equiv1,\quad a_3\equiv1,\quad a_5\equiv4,\quad a_7\equiv3,\quad a_9\equiv3,\quad a_{11}\equiv2 \pmod5.$$
None of these residues equals $0$. Since every odd index is congruent modulo $12$ to one of
$$1,3,5,7,9,11,$$
and the sequence is periodic modulo $5$ with period $12$, no odd-indexed term $a_m$ is divisible by $5$. ∎
Certification: this checks every odd residue class modulo the period; omitting one class would leave open the possibility of a hidden zero.
By Lemma 1,
$$S_n=a_{2n+1}.$$
The index $2n+1$ is odd. Lemma 4 shows that no odd-indexed term $a_m$ is congruent to $0\pmod5$. Hence
$$S_n\not\equiv0\pmod5$$
for every $n\ge0$.
Thus $S_n$ is never divisible by $5$.
This completes the proof.
∎
Verification of Key Steps
The first delicate step is the identity
$$S_n=a_{2n+1}.$$
Starting from
$$(1+\sqrt8)^{2n+1}-(1-\sqrt8)^{2n+1},$$
the coefficient of $(\sqrt8)^j$ becomes
$$1-(-1)^j.$$
For even $j$ this equals $0$, and for odd $j=2k+1$ it equals $2$. Hence the difference equals
$$2\sum_{k=0}^{n} \binom{2n+1}{2k+1} (\sqrt8)^{2k+1}.$$
Dividing by $2\sqrt8$ yields exactly the required sum. A careless computation could forget the factor $2$, producing an incorrect normalization.
The second delicate step is the recurrence. Since
$$(1+\sqrt8)^2=2(1+\sqrt8)+7,$$
multiplying by $(1+\sqrt8)^m$ gives
$$(1+\sqrt8)^{m+2} = 2(1+\sqrt8)^{m+1} + 7(1+\sqrt8)^m.$$
The same holds for $1-\sqrt8$. Subtracting yields the recurrence. An incorrect characteristic polynomial would destroy the entire modular argument.
The third delicate step is periodicity. The recurrence modulo $5$ is second order, so the state is the ordered pair
$$(a_m,a_{m+1}).$$
After computation,
$$(a_{12},a_{13})=(0,1)=(a_0,a_1).$$
Since the recurrence determines all future terms uniquely from a state, the sequence repeats. Matching only $a_{12}=a_0$ would not justify periodicity.
Alternative Approaches
A different approach starts directly from
$$S_n=\frac{(1+\sqrt8)^{2n+1}-(1-\sqrt8)^{2n+1}}{2\sqrt8}.$$
Working in the ring $\mathbb Z[\sqrt8]$ modulo $5$, one has
$$8\equiv3\pmod5,$$
and
$$(1+\sqrt8)^6\equiv-(1-\sqrt8)^6.$$
From this relation one can derive
$$S_{n+6}\equiv -S_n \pmod5,$$
which immediately yields a period dividing $12$. Checking six initial values then completes the proof.
Another possibility is to use the recurrence from the start and analyze the state space modulo $5$. Since there are only $25$ possible ordered pairs $(a_m,a_{m+1})$, every sequence is eventually periodic. Computing the orbit beginning with $(0,1)$ shows that it returns to the initial state after $12$ steps and that none of the odd-indexed terms in the cycle vanishes. The characteristic-root method used in the main proof is preferable because it explains naturally where the recurrence comes from and links it directly to the original binomial sum.