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Polish MO Finals
1976 Polish MO Finals
2
a_{n+1} = a_n +b_n 4 recurrence relations
a_{n+1} = a_n +b_n 4 recurrence relations
Source: Polish MO Finals 1976 p2
August 23, 2024
algebra
recurrence relation
Problem Statement
Four sequences of real numbers
(
a
n
)
,
(
b
n
)
,
(
c
n
)
,
(
d
n
)
(a_n), (b_n), (c_n), (d_n)
(
a
n
)
,
(
b
n
)
,
(
c
n
)
,
(
d
n
)
satisfy for all
n
n
n
,
a
n
+
1
=
a
n
+
b
n
,
b
n
+
1
=
b
n
+
c
n
,
a_{n+1} = a_n +b_n, b_{n+1} = b_n +c_n,
a
n
+
1
=
a
n
+
b
n
,
b
n
+
1
=
b
n
+
c
n
,
c
n
+
1
=
c
n
+
d
n
,
d
n
+
1
=
d
n
+
a
n
.
c_{n+1} = c_n +d_n, d_{n+1} = d_n +a_n.
c
n
+
1
=
c
n
+
d
n
,
d
n
+
1
=
d
n
+
a
n
.
Prove that if
a
k
+
m
=
a
m
,
b
k
+
m
=
b
m
,
c
k
+
m
=
c
m
,
d
k
+
m
=
d
m
a_{k+m} = a_m, b_{k+m} = b_m, c_{k+m} = c_m, d_{k+m} = d_m
a
k
+
m
=
a
m
,
b
k
+
m
=
b
m
,
c
k
+
m
=
c
m
,
d
k
+
m
=
d
m
for some
k
≥
1
,
n
≥
1
k\ge 1,n \ge 1
k
≥
1
,
n
≥
1
, then
a
2
=
b
2
=
c
2
=
d
2
=
0
a_2 = b_2 = c_2 = d_2 = 0
a
2
=
b
2
=
c
2
=
d
2
=
0
.
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