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2004 Italy TST
3
Functional equation with powers of 2
Functional equation with powers of 2
Source:
November 21, 2010
function
quadratics
algorithm
induction
floor function
algebra proposed
algebra
Problem Statement
Find all functions
f
:
N
→
N
f:\mathbb{N}\rightarrow \mathbb{N}
f
:
N
→
N
such that for all
m
,
n
∈
N
m,n\in\mathbb{N}
m
,
n
∈
N
,
(
2
m
+
1
)
f
(
n
)
f
(
2
m
n
)
=
2
m
f
(
n
)
2
+
f
(
2
m
n
)
2
+
(
2
m
−
1
)
2
n
.
(2^m+1)f(n)f(2^mn)=2^mf(n)^2+f(2^mn)^2+(2^m-1)^2n.
(
2
m
+
1
)
f
(
n
)
f
(
2
m
n
)
=
2
m
f
(
n
)
2
+
f
(
2
m
n
)
2
+
(
2
m
−
1
)
2
n
.
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