IMO 1974 Problem 3
For $n=0$ the sum equals
Proposed by: -
Verified: yes
Verdicts: PASS + PASS
Solve time: 5m13s
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
For $n=0$ the sum equals
$$\binom11 2^0=1,$$
which is not divisible by $5$.
For $n=1$,
$$\binom31 2^0+\binom33 2^3=3+8=11,$$
and $11\not\equiv0\pmod5$.
For $n=2$,
$$\binom51+\binom53 2^3+\binom55 2^6 =5+10\cdot8+64 =149,$$
and
$$149\equiv4\pmod5.$$
For $n=3$,
$$\binom71+\binom73 2^3+\binom75 2^6+\binom77 2^9 =7+35\cdot8+21\cdot64+512 =2143,$$
and
$$2143\equiv3\pmod5.$$
The values modulo $5$ for the first four cases are
$$1,1,4,3.$$
No obvious periodic pattern appears immediately.
The summation contains only odd binomial coefficients. A natural idea is to compare it with the expansion of
$$(1+x)^{2n+1}-(1-x)^{2n+1},$$
because the odd powers survive in the difference. Setting $x$ appropriately may convert the sum into a compact closed form.
Indeed,
$$(1+x)^{2n+1}-(1-x)^{2n+1} = 2\sum_{k=0}^n \binom{2n+1}{2k+1}x^{2k+1}.$$
If one chooses $x=2\sqrt2$, then
$$x^{2k+1}=2^{3k+\frac32},$$
which nearly matches the factor $2^{3k}$. This suggests dividing by a suitable power of $2$.
A cleaner substitution is obtained by rewriting
$$2^{3k}=8^k.$$
Then
$$\sum_{k=0}^n \binom{2n+1}{2k+1}8^k = \frac1{2\sqrt8}\left((1+\sqrt8)^{2n+1}-(1-\sqrt8)^{2n+1}\right).$$
Since $\sqrt8=2\sqrt2$, the denominator becomes $4\sqrt2$.
The next difficulty is reduction modulo $5$. Directly handling square roots modulo $5$ is awkward. A different path is needed.
Since
$$8\equiv3\pmod5,$$
the sum is congruent modulo $5$ to
$$\sum_{k=0}^n \binom{2n+1}{2k+1}3^k.$$
The closed form becomes
$$\frac1{2\sqrt3}\left((1+\sqrt3)^{2n+1}-(1-\sqrt3)^{2n+1}\right).$$
This still involves irrational quantities.
Another approach is to work directly with complex numbers. Because
$$1+2i$$
has norm $5$, powers of $1+2i$ naturally interact with divisibility by $5$. Expanding
$$(1+2i)^{2n+1}+(1-2i)^{2n+1}$$
or their difference isolates odd terms with powers of $-4$. After simplification, the given sum may appear.
Trying this,
$$(1+2i)^{2n+1}-(1-2i)^{2n+1} = 2i\sum_{k=0}^n \binom{2n+1}{2k+1}(-4)^k.$$
Modulo $5$,
$$(-4)^k\equiv1^k\equiv1.$$
This does not match the factor $8^k$.
A more promising idea is to derive a recurrence for the sum. Define
$$S_n=\sum_{k=0}^n \binom{2n+1}{2k+1}8^k.$$
Using the closed form, one expects a second-order linear recurrence. Computing the first values,
$$S_0=1,\quad S_1=11,\quad S_2=149,\quad S_3=2143.$$
Checking,
$$149=18\cdot11-49\cdot1,$$
$$2143=18\cdot149-49\cdot11.$$
Thus
$$S_n=18S_{n-1}-49S_{n-2}.$$
Modulo $5$ this becomes
$$S_n\equiv3S_{n-1}+S_{n-2}\pmod5.$$
Starting from
$$1,1,$$
the sequence modulo $5$ is
$$1,1,4,3,3,2,4,4,1,2,\dots$$
and none of these are $0$. The sequence seems periodic with period dividing $24$.
Still, a direct closed form modulo $5$ would be cleaner. Using the binomial identity with $x=2\sqrt2$,
$$S_n = \frac{(1+2\sqrt2)^{2n+1}-(1-2\sqrt2)^{2n+1}}{4\sqrt2}.$$
Let
$$a_n=(1+2\sqrt2)^n+(1-2\sqrt2)^n.$$
Then $a_n$ satisfies
$$a_n=2a_{n-1}+7a_{n-2}.$$
Modulo $5$,
$$a_n\equiv2a_{n-1}+2a_{n-2}.$$
This still requires connecting $a_n$ to $S_n$.
The most efficient route appears to be the recurrence for $S_n$ together with a complete modular analysis. The delicate point is proving rigorously that the recurrence indeed holds.
Problem Understanding
The problem asks for a proof that the integer
$$\sum_{k=0}^n \binom{2n+1}{2k+1}2^{3k}$$
is never divisible by $5$, for every integer $n\ge0$.
This is a Type B problem. A statement is given, and the task is to prove it for all nonnegative integers $n$.
The mathematical objects involved are binomial coefficients and powers of $2$. The sum ranges only over odd binomial indices. The main difficulty is that direct modular reduction does not immediately simplify the expression. The odd-index restriction prevents a straightforward application of the binomial theorem in its standard form.
The core idea is to transform the sum into a closed algebraic expression using
$$(1+x)^{2n+1}-(1-x)^{2n+1},$$
which isolates odd powers. From that closed form one obtains a linear recurrence relation for the sequence. The recurrence modulo $5$ becomes periodic, and one can prove that the residue $0$ never occurs.
Proof Architecture
Define
$$S_n=\sum_{k=0}^n \binom{2n+1}{2k+1}8^k.$$
The first lemma expresses $S_n$ in closed form:
$$S_n=\frac{(1+2\sqrt2)^{2n+1}-(1-2\sqrt2)^{2n+1}}{4\sqrt2}.$$
This follows from the identity obtained by subtracting the binomial expansions of $(1+x)^{2n+1}$ and $(1-x)^{2n+1}$.
The second lemma proves that $(S_n)$ satisfies the recurrence
$$S_n=18S_{n-1}-49S_{n-2}$$
for every $n\ge2$. This is derived from the fact that the two numbers
$$(1+2\sqrt2)^2,\qquad (1-2\sqrt2)^2$$
are the roots of
$$t^2-18t+49=0.$$
The third lemma studies the sequence modulo $5$. Reducing the recurrence modulo $5$ gives
$$S_n\equiv3S_{n-1}+S_{n-2}\pmod5.$$
Using the initial values
$$S_0=1,\qquad S_1=11\equiv1\pmod5,$$
one computes enough consecutive terms to establish a period and verify that none are congruent to $0$ modulo $5$.
The most delicate step is the derivation of the recurrence from the closed form. A careless argument may manipulate irrational expressions formally without justifying why the resulting recurrence applies to the integer sequence itself.
Solution
Define
$$S_n=\sum_{k=0}^n \binom{2n+1}{2k+1}8^k.$$
The required statement is that
$$5\nmid S_n$$
for every integer $n\ge0$.
Lemma 1
For every integer $n\ge0$,
$$S_n= \frac{(1+2\sqrt2)^{2n+1}-(1-2\sqrt2)^{2n+1}}{4\sqrt2}.$$
Proof
For any real number $x$,
$$(1+x)^{2n+1} = \sum_{j=0}^{2n+1}\binom{2n+1}{j}x^j,$$
and
$$(1-x)^{2n+1} = \sum_{j=0}^{2n+1}\binom{2n+1}{j}(-x)^j.$$
Subtracting the second identity from the first gives
$$(1+x)^{2n+1}-(1-x)^{2n+1} = 2\sum_{k=0}^n \binom{2n+1}{2k+1}x^{2k+1}.$$
Choose
$$x=2\sqrt2.$$
Then
$$x^{2k+1} = (2\sqrt2)^{2k+1} = 2^{3k+1}\sqrt2.$$
Hence
$$(1+2\sqrt2)^{2n+1}-(1-2\sqrt2)^{2n+1} = 2\sum_{k=0}^n \binom{2n+1}{2k+1}2^{3k+1}\sqrt2.$$
Factoring out $4\sqrt2$ yields
$$(1+2\sqrt2)^{2n+1}-(1-2\sqrt2)^{2n+1} = 4\sqrt2,S_n.$$
Therefore
$$S_n= \frac{(1+2\sqrt2)^{2n+1}-(1-2\sqrt2)^{2n+1}}{4\sqrt2}.$$
∎
This lemma converts the combinatorial sum into an algebraic expression; omitting the explicit computation of $(2\sqrt2)^{2k+1}$ would risk an incorrect power of $2$.
Lemma 2
For every integer $n\ge2$,
$$S_n=18S_{n-1}-49S_{n-2}.$$
Proof
Set
$$\alpha=(1+2\sqrt2)^2, \qquad \beta=(1-2\sqrt2)^2.$$
Direct computation gives
$$\alpha=9+4\sqrt2, \qquad \beta=9-4\sqrt2.$$
Hence
$$\alpha+\beta=18, \qquad \alpha\beta=49.$$
Therefore $\alpha$ and $\beta$ are the two roots of
$$t^2-18t+49=0.$$
Consequently,
$$\alpha^2=18\alpha-49, \qquad \beta^2=18\beta-49.$$
By Lemma 1,
$$S_n= \frac{(1+2\sqrt2)\alpha^n-(1-2\sqrt2)\beta^n}{4\sqrt2}.$$
For $n\ge2$,
$$\begin{aligned} 18S_{n-1}-49S_{n-2} &= \frac{18(1+2\sqrt2)\alpha^{n-1}-18(1-2\sqrt2)\beta^{n-1}}{4\sqrt2} \ &\quad- \frac{49(1+2\sqrt2)\alpha^{n-2}-49(1-2\sqrt2)\beta^{n-2}}{4\sqrt2}. \end{aligned}$$
Combining the fractions,
$$\begin{aligned} 18S_{n-1}-49S_{n-2} &= \frac{(1+2\sqrt2)\alpha^{n-2}(18\alpha-49)}{4\sqrt2} \ &\quad- \frac{(1-2\sqrt2)\beta^{n-2}(18\beta-49)}{4\sqrt2}. \end{aligned}$$
Using
$$18\alpha-49=\alpha^2, \qquad 18\beta-49=\beta^2,$$
we obtain
$$\begin{aligned} 18S_{n-1}-49S_{n-2} &= \frac{(1+2\sqrt2)\alpha^n-(1-2\sqrt2)\beta^n}{4\sqrt2} \ &= S_n. \end{aligned}$$
Thus
$$S_n=18S_{n-1}-49S_{n-2}$$
for all $n\ge2$.
∎
This lemma establishes a recurrence for the integer sequence itself; skipping the substitution of the quadratic identities for $\alpha$ and $\beta$ would leave the recurrence unproved.
Lemma 3
For every integer $n\ge0$,
$$S_n\not\equiv0\pmod5.$$
Proof
Reducing the recurrence of Lemma 2 modulo $5$ gives
$$S_n\equiv3S_{n-1}+S_{n-2}\pmod5$$
for all $n\ge2$, since
$$18\equiv3\pmod5, \qquad -49\equiv1\pmod5.$$
The initial values are
$$S_0=\binom11=1,$$
and
$$S_1=\binom31+\binom338 =3+8 =11 \equiv1\pmod5.$$
We now compute successive residues modulo $5$ using the recurrence:
$$S_2\equiv3\cdot1+1\equiv4,$$
$$S_3\equiv3\cdot4+1\equiv13\equiv3,$$
$$S_4\equiv3\cdot3+4\equiv13\equiv3,$$
$$S_5\equiv3\cdot3+3\equiv12\equiv2,$$
$$S_6\equiv3\cdot2+3\equiv9\equiv4,$$
$$S_7\equiv3\cdot4+2\equiv14\equiv4,$$
$$S_8\equiv3\cdot4+4\equiv16\equiv1,$$
$$S_9\equiv3\cdot1+4\equiv7\equiv2,$$
$$S_{10}\equiv3\cdot2+1\equiv7\equiv2,$$
$$S_{11}\equiv3\cdot2+2\equiv8\equiv3,$$
$$S_{12}\equiv3\cdot3+2\equiv11\equiv1,$$
$$S_{13}\equiv3\cdot1+3\equiv6\equiv1.$$
The ordered pair
$$(S_{12},S_{13})$$
equals
$$(1,1),$$
which is the same as
$$(S_0,S_1).$$
Since each subsequent term is uniquely determined by the previous two terms through the recurrence
$$S_n\equiv3S_{n-1}+S_{n-2}\pmod5,$$
the sequence modulo $5$ is periodic with period $12$.
Among the residues
$$1,1,4,3,3,2,4,4,1,2,2,3$$
appearing in one complete period, none equals $0$. Hence
$$S_n\not\equiv0\pmod5$$
for every integer $n\ge0$.
∎
This lemma finishes the modular analysis; merely checking finitely many initial values without proving periodicity would not justify the statement for all $n$.
The number
$$\sum_{k=0}^n \binom{2n+1}{2k+1}2^{3k}$$
is therefore not divisible by $5$ for any integer $n\ge0$.
This completes the proof.
∎
Verification of Key Steps
The first delicate step is the identity
$$(1+x)^{2n+1}-(1-x)^{2n+1} = 2\sum_{k=0}^n\binom{2n+1}{2k+1}x^{2k+1}.$$
To verify it independently, expand both powers:
$$(1+x)^{2n+1} = \sum_{j=0}^{2n+1}\binom{2n+1}{j}x^j,$$
$$(1-x)^{2n+1} = \sum_{j=0}^{2n+1}\binom{2n+1}{j}(-1)^jx^j.$$
Subtracting gives coefficient
$$1-(-1)^j.$$
If $j$ is even, this coefficient equals $0$. If $j$ is odd, it equals $2$. Hence only odd powers remain. A careless derivation may forget the factor $2$, which would produce an incorrect final constant.
The second delicate step is the recurrence
$$S_n=18S_{n-1}-49S_{n-2}.$$
An independent derivation starts from
$$(1+2\sqrt2)^2=9+4\sqrt2,$$
$$(1-2\sqrt2)^2=9-4\sqrt2.$$
Their sum is $18$ and product is $49$, so both satisfy
$$u^2=18u-49.$$
If
$$T_n=A\alpha^n+B\beta^n,$$
then
$$T_n = A\alpha^{n-2}\alpha^2+B\beta^{n-2}\beta^2 = 18T_{n-1}-49T_{n-2}.$$
The error most likely to occur here is using $\alpha+\beta$ correctly but computing $\alpha\beta$ incorrectly. Since
$$(9+4\sqrt2)(9-4\sqrt2)=81-32=49,$$
the coefficient really is $49$.
The third delicate step is the periodicity modulo $5$. The recurrence determines every future pair
$$(S_n,S_{n+1})$$
from the current pair. Once a pair repeats, the entire subsequent sequence repeats identically. Since
$$(S_{12},S_{13})=(1,1)=(S_0,S_1),$$
the sequence from index $12$ onward matches the sequence from index $0$ onward. A careless argument may observe a repeated single value rather than a repeated ordered pair, which does not guarantee periodicity for a second-order recurrence.
Alternative Approaches
A different proof uses arithmetic in the quadratic ring $\mathbb{Z}[\sqrt2]$. From Lemma 1,
$$S_n= \frac{(1+2\sqrt2)^{2n+1}-(1-2\sqrt2)^{2n+1}}{4\sqrt2}.$$
Modulo $5$, the element $2$ is a quadratic nonresidue, so adjoining $\sqrt2$ produces the finite field with $25$ elements. In that field one has
$$(1+2\sqrt2)^5=1-2\sqrt2,$$
which leads to a periodicity relation for the numerator. After simplifying, one finds directly that $S_n$ cycles through nonzero residues modulo $5$.
Another approach derives the recurrence combinatorially instead of through radicals. Starting from
$$S_n=\sum_{k=0}^n \binom{2n+1}{2k+1}8^k,$$
one may use Pascal identities repeatedly to express $S_n$ in terms of $S_{n-1}$ and $S_{n-2}$. This avoids irrational numbers entirely, but the algebra becomes considerably longer and obscures the origin of the coefficients $18$ and $49$. The algebraic closed-form method isolates the structure immediately and leads to the recurrence systematically.