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Putnam
2019 Putnam
B5
B5
Part of
2019 Putnam
Problems
(1)
Putnam 2019 B5
Source:
12/10/2019
Let
F
m
F_m
F
m
be the
m
m
m
'th Fibonacci number, defined by
F
1
=
F
2
=
1
F_1=F_2=1
F
1
=
F
2
=
1
and
F
m
=
F
m
−
1
+
F
m
−
2
F_m = F_{m-1}+F_{m-2}
F
m
=
F
m
−
1
+
F
m
−
2
for all
m
≥
3
m \geq 3
m
≥
3
. Let
p
(
x
)
p(x)
p
(
x
)
be the polynomial of degree 1008 such that
p
(
2
n
+
1
)
=
F
2
n
+
1
p(2n+1)=F_{2n+1}
p
(
2
n
+
1
)
=
F
2
n
+
1
for
n
=
0
,
1
,
2
,
…
,
1008
n=0,1,2,\ldots,1008
n
=
0
,
1
,
2
,
…
,
1008
. Find integers
j
j
j
and
k
k
k
such that
p
(
2019
)
=
F
j
−
F
k
p(2019) = F_j - F_k
p
(
2019
)
=
F
j
−
F
k
.