51st International Mathematical Olympiad

Astana, Kazakhstan 2010

Problems with Solutions

Contents

Problems 5

Solutions 7
Problem 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
Problem 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
Problem 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
Problem 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
Problem 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
Problem 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
Problems

Problem 1. Determine all functions f : R → R such that the equality

( ) ⌊ ⌋
f ⌊x⌋y = f (x) f (y)
holds for all x, y ∈ R. (Here ⌊z⌋ denotes the greatest integer less than or equal to z.)

Problem 2. Let I be the incentre of triangle ABC and let Γ be its circumcircle. Let the line AI
\ and F a point on the side BC such that
intersect Γ again at D. Let E be a point on the arc BDC
∠BAF = ∠CAE < 12 ∠BAC.
Finally, let G be the midpoint of the segment IF . Prove that the lines DG and EI intersect on Γ.

Problem 3. Let N be the set of positive integers. Determine all functions g : N → N such that
( )( )
g(m) + n m + g(n)
is a perfect square for all m, n ∈ N.

Problem 4. Let P be a point inside the triangle ABC. The lines AP , BP and CP intersect the
circumcircle Γ of triangle ABC again at the points K, L and M respectively. The tangent to Γ at C
intersects the line AB at S. Suppose that SC = SP . Prove that M K = M L.

Problem 5. In each of six boxes B1 , B2 , B3 , B4 , B5 , B6 there is initially one coin. There are two
types of operation allowed:
Type 1: Choose a nonempty box Bj with 1 ≤ j ≤ 5. Remove one coin from Bj and add two
coins to Bj+1 .
Type 2: Choose a nonempty box Bk with 1 ≤ k ≤ 4. Remove one coin from Bk and exchange
the contents of (possibly empty) boxes Bk+1 and Bk+2 .
Determine whether there is a finite sequence of such operations that results in boxes B1 , B2 , B3 , B4 , B5
2010 c c
being empty and box B6 containing exactly 20102010 coins. (Note that ab = a(b ) .)

Problem 6. Let a1 , a2 , a3 , . . . be a sequence of positive real numbers. Suppose that for some
positive integer s, we have
an = max{ak + an−k | 1 ≤ k ≤ n − 1}
for all n > s. Prove that there exist positive integers ℓ and N , with ℓ ≤ s and such that an = aℓ +an−ℓ
for all n ≥ N .
6
Solutions

Problem 1. Determine all functions f : R → R such that the equality

( ) ⌊ ⌋
f ⌊x⌋y = f (x) f (y) (1)
holds for all x, y ∈ R. (Here ⌊z⌋ denotes the greatest integer less than or equal to z.)
Answer. f (x) = const = C, where C = 0 or 1 ≤ C < 2.
Solution 1. First, setting x = 0 in (1) we get
f (0) = f (0)⌊f (y)⌋ (2)
for all y ∈ R. Now, two cases are possible.
Case 1. Assume that f (0) ̸= 0. Then from (2) we conclude that ⌊f (y)⌋ = 1 for all y ∈ R.
Therefore, equation (1) becomes f (⌊x⌋y) = f (x), and substituting y = 0 we have f (x) = f (0) =
C ̸= 0. Finally, from ⌊f (y)⌋ = 1 = ⌊C⌋ we obtain that 1 ≤ C < 2.
Case 2. Now we have f (0) = 0. Here we consider two subcases.
Subcase 2a. Suppose that there exists 0 < α < 1 such that f (α) ̸= 0. Then setting x = α in (1)
we obtain 0 = f (0) = f (α)⌊f (y)⌋ for all y ∈ R. Hence, ⌊f (y)⌋ = 0 for all y ∈ R. Finally, substituting
x = 1 in (1) provides f (y) = 0 for all y ∈ R, thus contradicting the condition f (α) ̸= 0.
Subcase 2b. Conversely, we have f (α) = 0 for all 0 ≤ α < 1. Consider any real z; there exists an
z
integer N such that α = ∈ [0, 1) (one may set N = ⌊z⌋ + 1 if z ≥ 0 and N = ⌊z⌋ − 1 otherwise).
N
Now, from (1) we get f (z) = f (⌊N ⌋α) = f (N )⌊f (α)⌋ = 0 for all z ∈ R.
Finally, a straightforward check shows that all the obtained functions satisfy (1).
Solution 2. Assume that ⌊f (y)⌋ = 0 for some y; then the substitution x = 1 provides f (y) =
f (1)⌊f (y)⌋ = 0. Hence, if ⌊f (y)⌋ = 0 for all y, then f (y) = 0 for all y. This function obviously
satisfies the problem conditions.
So we are left to consider the case when ⌊f (a)⌋ ̸= 0 for some a. Then we have
f (⌊x⌋a)
f (⌊x⌋a) = f (x)⌊f (a)⌋, or f (x) = . (3)
⌊f (a)⌋
This means that f (x1 ) = f (x2 ) whenever ⌊x1 ⌋ = ⌊x2 ⌋, hence f (x) = f (⌊x⌋), and we may assume
that a is an integer.
Now we have ( ) ⌊ ( )⌋
f (a) = f 2a · 12 = f (2a) f 12 = f (2a)⌊f (0)⌋;
this implies ⌊f (0)⌋ ̸= 0, so we may even assume that a = 0. Therefore equation (3) provides
f (0)
f (x) = = C ̸= 0
⌊f (0)⌋
8

for each x. Now, condition (1) becomes equivalent to the equation C = C⌊C⌋ which holds exactly
when ⌊C⌋ = 1.

Problem 2. Let I be the incentre of triangle ABC and let Γ be its circumcircle. Let the line AI
\ and F a point on the side BC such that
intersect Γ again at D. Let E be a point on the arc BDC
∠BAF = ∠CAE < 12 ∠BAC.
Finally, let G be the midpoint of the segment IF . Prove that the lines DG and EI intersect on Γ.
Solution 1. Let X be the second point of intersection of line EI with Γ, and L be the foot of the
bisector of angle BAC. Let G′ and T be the points of intersection of segment DX with lines IF
and AF , respectively. We are to prove that G = G′ , or IG′ = G′ F . By the Menelaus theorem
applied to triangle AIF and line DX, it means that we need the relation
G′ F T F AD TF ID
1= ′
= · , or = .
IG AT ID AT AD
Let the line AF intersect Γ at point K ̸= A (see Fig. 1); since ∠BAK = ∠CAE we have
d
BK = CE,d hence KE ∥ BC. Notice that ∠IAT = ∠DAK = ∠EAD = ∠EXD = ∠IXT , so
the points I, A, X, T are concyclic. Hence we have ∠IT A = ∠IXA = ∠EXA = ∠EKA, so
TF IL
IT ∥ KE ∥ BC. Therefore we obtain = .
AT AI
IL CL
Since CI is the bisector of ∠ACL, we get = . Furthermore, ∠DCL = ∠DCB =
AI AC
∠DAB = ∠CAD = 12 ∠BAC, hence the triangles DCL and DAC are similar; therefore we get
CL DC
= . Finally, it is known that the midpoint D of arc BC is equidistant from points I, B, C,
AC AD
DC ID
hence = .
AD AD
Summarizing all these equalities, we get
TF IL CL DC ID
= = = = ,
AT AI AC AD AD
as desired.
X A
A

B C
III
TTT
G′
D
F
B L C

K E
D J
Fig. 1 Fig. 2
 9

AI AD
Comment. The equality = is known and can be obtained in many different ways. For instance,
IL DI
\ to the
one can consider the inversion with center D and radius DC = DI. This inversion takes BAC
IL AI
segment BC, so point A goes to L. Hence = , which is the desired equality.
DI AD
Solution 2. As in the previous solution, we introduce the points X, T and K and note that it
suffice to prove the equality
TF DI T F + AT DI + AD AT AF
= ⇐⇒ = ⇐⇒ = .
AT AD AT AD AD DI + AD
Since ∠F AD = ∠EAI and ∠T DA = ∠XDA = ∠XEA = ∠IEA, we get that the triangles AT D
AT AI
and AIE are similar, therefore = .
AD AE
Next, we also use the relation DB = DC = DI. Let J be the point on the extension of
segment AD over point D such that DJ = DI = DC (see Fig. 2). Then ∠DJC = ∠JCD =
1
2
(π − ∠JDC) = 12 ∠ADC = 12 ∠ABC = ∠ABI. Moreover, ∠BAI = ∠JAC, hence triangles ABI
AB AI
and AJC are similar, so = , or AB · AC = AJ · AI = (DI + AD) · AI.
AJ AC
On the other hand, we get ∠ABF = ∠ABC = ∠AEC and ∠BAF = ∠CAE, so triangles ABF
AF AB
and AEC are also similar, which implies = , or AB · AC = AF · AE.
AC AE
Summarizing we get
AI AF AT AF
(DI + AD) · AI = AB · AC = AF · AE ⇒ = ⇒ = ,
AE AD + DI AD AD + DI
as desired.
Comment. In fact, point J is an excenter of triangle ABC.

Problem 3. Let N be the set of positive integers. Determine all functions g : N → N such that
( )( )
g(m) + n m + g(n)
is a perfect square for all m, n ∈ N.
Answer. All functions of the form g(n) = n + c, where c ∈ N ∪ {0}.
Solution. First, it is clear that all functions of( the form)(g(n) = n + ) c with a constant nonnegative
2
integer c satisfy the problem conditions since g(m) + n g(n) + m = (n + m + c) is a square.
We are left to prove that there are no other functions. We start with the following
 
Lemma. Suppose that p  g(k) − g(ℓ) for some prime p and positive integers k, ℓ. Then p  k − ℓ.

Proof. Suppose first that p2  g(k) − g(ℓ), so g(ℓ) = g(k) + p2 a for some integer a. Take some positive
integer D > max{g(k), g(ℓ)} which is not divisible( ) set n = pD − g(k). Then the positive
by p and
numbers n + g(k) = pD and n + g(ℓ) = pD + g(ℓ) − g(k) = p(D + pa) are both ( divisible
)( by p but)
2
not by
( p . Now,
)( applying
) the problem conditions, we get that both the numbers g(k) + n g(n) + k
2
and g(ℓ) + n g(n) + ℓ are squares divisible by p (and thus  ( by p );)this( means that) the multipliers
g(n) + k and g(n) + ℓ are also divisible by p, therefore p  g(n) + k − g(n) + ℓ = k − ℓ as well.
On the other hand, if g(k)−g(ℓ) is divisible by p but not by p2 , then choose the same ( number D)and
set n = p D −g(k). Then the positive numbers g(k)+n = p D and g(ℓ)+n = p D + g(ℓ)−g(k) are
3 3 3

respectively divisible by p3 (but not by p4 ) and by p (but not by p2 ). Hence

 ( in analogous
) ( way we ) obtain

that the numbers g(n) + k and g(n) + ℓ are divisible by p, therefore p g(n) + k − g(n) + ℓ = k − ℓ.

10

We turn to the problem. First, suppose that g(k) = g(ℓ) for some k, ℓ ∈ N. Then by Lemma we
have that k − ℓ is divisible by every prime number, so k − ℓ = 0, or k = ℓ. Therefore, the function g
is injective.
Next, consider the numbers g(k) and g(k + 1). Since the number (k + 1) − k = 1 has no prime
divisors, by Lemma the same holds for g(k + 1) − g(k); thus |g(k + 1) − g(k)| = 1.
Now, let g(2) − g(1) = q, |q| = 1. Then we prove by induction that g(n) = g(1) + q(n − 1). The
base for n = 1, 2 holds by the definition of q. For the step, if n > 1 we have g(n + 1) = g(n) ± q =
g(1) + q(n − 1) ± q. Since g(n) ̸= g(n − 2) = g(1) + q(n − 2), we get g(n) = g(1) + qn, as desired.
Finally, we have g(n) = g(1) + q(n − 1). Then q cannot be −1 since otherwise for n ≥ g(1) + 1
we have g(n) ≤ 0 which is impossible. Hence q = 1 and g(n) = (g(1) − 1) + n for each n ∈ N, and
g(1) − 1 ≥ 0, as desired.

Problem 4. Let P be a point inside the triangle ABC. The lines AP , BP and CP intersect the
circumcircle Γ of triangle ABC again at the points K, L and M respectively. The tangent to Γ at C
intersects the line AB at S. Suppose that SC = SP . Prove that M K = M L.
Solution 1. We assume that CA > CB, so point S lies on the ray AB.
PM PA
From the similar triangles △P KM ∼ △P CA and △P LM ∼ △P CB we get = and
KM CA
LM CB
= . Multiplying these two equalities, we get
PM PB
LM CB P A
= · .
KM CA P B
CB PB
Hence, the relation M K = M L is equivalent to = .
CA PA
Denote by E the foot of the bisector of angle B in triangle ABC. Recall that the locus of points X
XA CA
for which = is the Apollonius circle Ω with the center Q on the line AB, and this circle
XB CB
passes through C and E. Hence, we have M K = M L if and only if P lies on Ω, that is QP = QC.

Ω
L
C
C
C

P
P
P

S
A E B

M
Fig. 1
 11

Now we prove that S = Q, thus establishing the problem statement. We have ∠CES = ∠CAE +
∠ACE = ∠BCS + ∠ECB = ∠ECS, so SC = SE. Hence, the point S lies on AB as well as on the
perpendicular bisector of CE and therefore coincides with Q.
Comment. In this solution we proved more general fact: SC = SP if and only if M K = M L.
Solution 2. As in the previous solution, we assume that S lies on the ray AB.
Let P be an arbitrary point inside both the circumcircle ω of the triangle ABC and the angle
ASC, the points K, L, M defined as in the problem.
Let E and F be the points of intersection of the line SP with ω, point E lying on the segment SP
(see Fig. 2).

F L

K
ω
C
P
P
P

M
E

A B S
Fig. 2

SP SA
We have SP 2 = SC 2 = SA · SB, so= , and hence △P SA ∼ △BSP . Then ∠BP S =
SB SP
d + LF
∠SAP . Since 2∠BP S = BE c and 2∠SAP = BE
d + EKd we have

c = EK.
LF d (4)

d+M
On the other hand, from ∠SP C = ∠SCP we have EC d d + EM
F = EC d , or

d
M d.
F = EM (5)

\
From (4) and (5) we get M d
FL = M F + FcL = M
d d =M
E + EK \ EK and hence M K = M L. The
claim is proved.

Problem 5. In each of six boxes B1 , B2 , B3 , B4 , B5 , B6 there is initially one coin. There are two
types of operation allowed:

Type 1: Choose a nonempty box Bj with 1 ≤ j ≤ 5. Remove one coin from Bj and add two
coins to Bj+1 .
Type 2: Choose a nonempty box Bk with 1 ≤ k ≤ 4. Remove one coin from Bk and exchange
the contents of (possibly empty) boxes Bk+1 and Bk+2 .
12

Determine whether there is a finite sequence of such operations that results in boxes B1 , B2 , B3 , B4 , B5
2010 c c
being empty and box B6 containing exactly 20102010 coins. (Note that ab = a(b ) .)

Answer. Yes. There exists such a sequence of moves.

Solution. Denote by (a1 , a2 , . . . , an ) → (a′1 , a′2 , . . . , a′n ) the following: if some consecutive boxes
contain a1 , . . . , an coins, then it is possible to perform several allowed moves such that the boxes
contain a′1 , . . . , a′n coins respectively, whereas the contents of the other boxes remain unchanged.
2010
Let A = 20102010 , respectively. Our goal is to show that

(1, 1, 1, 1, 1, 1) → (0, 0, 0, 0, 0, A).

First we prove two auxiliary observations.

Lemma 1. (a, 0, 0) → (0, 2a , 0) for every a ≥ 1.
Proof. We prove by induction that (a, 0, 0) → (a − k, 2k , 0) for every 1 ≤ k ≤ a. For k = 1, apply
Type 1 to the first box:
(a, 0, 0) → (a − 1, 2, 0) = (a − 1, 21 , 0).

Now assume that k < a and the statement holds for some k < a. Starting from (a − k, 2k , 0),
apply Type 1 to the middle box 2k times, until it becomes empty. Then apply Type 2 to the first
box:
(a − k, 2k , 0) → (a − k, 2k − 1, 2) → · · · → (a − k, 0, 2k+1 ) → (a − k − 1, 2k+1 , 0).

Hence,
(a, 0, 0) → (a − k, 2k , 0) → (a − k − 1, 2k+1 , 0). 

2
..
2.
(e.g. P3 = 22 = 16). Then (a, 0, 0, 0) →
2
Lemma 2. For every positive integer n, let Pn = |{z}
2
n
(0, Pa , 0, 0) for every a ≥ 1.
Proof. Similarly to Lemma 1, we prove that (a, 0, 0, 0) → (a − k, Pk , 0, 0) for every 1 ≤ k ≤ a.
For k = 1, apply Type 1 to the first box:

(a, 0, 0, 0) → (a − 1, 2, 0, 0) = (a − 1, P1 , 0, 0).

Now assume that the lemma holds for some k < a. Starting from (a − k, Pk , 0, 0), apply Lemma 1,
then apply Type 1 to the first box:

(a − k, Pk , 0, 0) → (a − k, 0, 2Pk , 0) = (a − k, 0, Pk+1 , 0) → (a − k − 1, Pk+1 , 0, 0).

Therefore,
(a, 0, 0, 0) → (a − k, Pk , 0, 0) → (a − k − 1, Pk+1 , 0, 0). 
 13

Now we prove the statement of the problem.

First apply Type 1 to box 5, then apply Type 2 to boxes B4 , B3 , B2 and B1 in this order. Then
apply Lemma 2 twice:

(1, 1, 1, 1, 1, 1) → (1, 1, 1, 1, 0, 3) → (1, 1, 1, 0, 3, 0) → (1, 1, 0, 3, 0, 0) → (1, 0, 3, 0, 0, 0) →

→ (0, 3, 0, 0, 0, 0) → (0, 0, P3 , 0, 0, 0) = (0, 0, 16, 0, 0, 0) → (0, 0, 0, P16 , 0, 0).

We already have more than A coins in box B4 , since

2010 2010 2010 2011 11 )2011 11·2011 215
A ≤ 20102010 < (211 )2010 = 211·2010 < 22010 < 2(2 = 22 < 22 < P16 .

To decrease the number of coins in box B4 , apply Type 2 to this stack repeatedly until its size
decreases to A/4. (In every step, we remove a coin from B4 and exchange the empty boxes B5
and B6 .)

(0, 0, 0, P16 , 0, 0) → (0, 0, 0, P16 − 1, 0, 0) → (0, 0, 0, P16 − 2, 0, 0) →

→ · · · → (0, 0, 0, A/4, 0, 0).

Finally, apply Type 1 repeatedly to empty boxes B4 and B5 :

(0, 0, 0, A/4, 0, 0) → · · · → (0, 0, 0, 0, A/2, 0) → · · · → (0, 0, 0, 0, 0, A).

Comment. Starting with only 4 boxes, it is not hard to check manually that we can achieve at most 28
coins in the last position. However, around 5 and 6 boxes the maximal number of coins explodes. With 5
14
boxes it is possible to achieve more than 22 coins. With 6 boxes the maximum is greater than PP 14 .
2

Problem 6. Let a1 , a2 , a3 , . . . be a sequence of positive real numbers. Suppose that for some
positive integer s, we have
an = max{ak + an−k | 1 ≤ k ≤ n − 1} (6)
for all n > s. Prove that there exist positive integers ℓ and N , with ℓ ≤ s and such that an = aℓ +an−ℓ
for all n ≥ N .
Solution 1. First, from the problem conditions we have that each an (n > s) can be expressed as
an = aj1 + aj2 with j1 , j2 < n, j1 + j2 = n. If, say, j1 > s then we can proceed in the same way
with aj1 , and so on. Finally, we represent an in a form

an = ai1 + · · · + aik , (7)

1 ≤ ij ≤ s, i1 + · · · + ik = n. (8)

Moreover, if ai1 and ai2 are the numbers in (7) obtained on the last step, then i1 + i2 > s. Hence we
can adjust (8) as
1 ≤ ij ≤ s, i1 + · · · + ik = n, i1 + i2 > s. (9)
On the other hand, suppose that the indices i1 , . . . , ik satisfy the conditions (9). Then, denoting
sj = i1 + · · · + ij , from (6) we have

an = ask ≥ ask−1 + aik ≥ ask−2 + aik−1 + aik ≥ · · · ≥ ai1 + · · · + aik .

Summarizing these observations we get the following

14

Claim. For every n > s, we have

an = max{ai1 + · · · + aik : the collection (i1 , . . . , ik ) satisfies (9)}. 

Now we denote
ai
m = max
1≤i≤s i
aℓ
and fix some index ℓ ≤ s such that m = .
ℓ
Consider some n ≥ s2 ℓ + 2s and choose an expansion of an in the form (7), (9). Then we have
n = i1 + · · · + ik ≤ sk, so k ≥ n/s ≥ sℓ + 2. Suppose that none of the numbers i3 , . . . , ik equals ℓ.
Then by the pigeonhole principle there is an index 1 ≤ j ≤ s which appears among i3 , . . . , ik at
least ℓ times, and surely j ̸= ℓ. Let us delete these ℓ occurrences of j from (i1 , . . . , ik ), and add
j occurrences of ℓ instead, obtaining a sequence (i1 , i2 , i′3 , . . . , i′k′ ) also satisfying (9). By Claim, we
have
ai1 + · · · + aik = an ≥ ai1 + ai2 + ai′3 + · · · + ai′k′ ,
aℓ aj
or, after removing the coinciding terms, ℓaj ≥ jaℓ , so ≤ . By the definition of ℓ, this means
ℓ j
that ℓaj = jaℓ , hence
an = ai1 + ai2 + ai′3 + · · · + ai′k′ .
Thus, for every n ≥ s2 ℓ + 2s we have found a representation of the form (7), (9) with ij = ℓ for
some j ≥ 3. Rearranging the indices we may assume that ik = ℓ.
Finally, observe that in this representation, the indices (i1 , . . . , ik−1 ) satisfy the conditions (9)
with n replaced by n − ℓ. Thus, from the Claim we get

an−ℓ + aℓ ≥ (ai1 + · · · + aik−1 ) + aℓ = an ,

which by (6) implies

an = an−ℓ + aℓ for each n ≥ s2 ℓ + 2s,
as desired.
Solution 2. As in the previous solution, we involve the expansion (7), (8), and we fix some index
1 ≤ ℓ ≤ s such that
aℓ ai
= m = max .
ℓ 1≤i≤s i

Now, we introduce the sequence (bn ) as bn = an − mn; then bℓ = 0.

We prove by induction on n that bn ≤ 0, and (bn ) satisfies the same recurrence relation as (an ).
The base cases n ≤ s follow from the definition of m. Now, for n > s from the induction hypothesis
we have

bn = max (ak + an−k ) − nm = max (bk + bn−k + nm) − nm = max (bk + bn−k ) ≤ 0,
1≤k≤n−1 1≤k≤n−1 1≤k≤n−1

as required.
Now, if bk = 0 for all 1 ≤ k ≤ s, then bn = 0 for all n, hence an = mn, and the statement is
trivial. Otherwise, define

M = max |bi |, ε = min{|bi | : 1 ≤ i ≤ s, bi < 0}.

1≤i≤s
 15

Then for n > s we obtain

bn = max (bk + bn−k ) ≥ bℓ + bn−ℓ = bn−ℓ ,

1≤k≤n−1

so
0 ≥ bn ≥ bn−ℓ ≥ bn−2ℓ ≥ · · · ≥ −M.
Thus, in view of the expansion (7), (8) applied to the sequence (bn ), we get that each bn is
contained in a set
T = {bi1 + bi2 + · · · + bik : i1 , . . . , ik ≤ s} ∩ [−M, 0]
We claim that this set is finite. Actually, for any x ∈ T , let x = bi1 + · · · + bik (i1 , . . . , ik ≤ s). Then
M M
among bij ’s there are at most nonzero terms (otherwise x < · (−ε) < −M ). Thus x can be
ε ε
M
expressed in the same way with k ≤ , and there is only a finite number of such sums.
ε
Finally, for every t = 1, 2, . . . , ℓ we get that the sequence

bs+t , bs+t+ℓ , bs+t+2ℓ , . . .

is non-decreasing and attains the finite number of values; therefore it is constant from some index.
Thus, the sequence (bn ) is periodic with period ℓ from some index N , which means that

bn = bn−ℓ = bn−ℓ + bℓ for all n > N + ℓ,

and hence

an = bn + nm = (bn−ℓ + (n − ℓ)m) + (bℓ + ℓm) = an−ℓ + aℓ for all n > N + ℓ,

as desired.