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National and Regional Contests
Netherlands Contests
Dutch BxMO/EGMO TST
2019 Dutch BxMO TST
1
1
Part of
2019 Dutch BxMO TST
Problems
(1)
a^2 + b = n, a+b = 2^k for each positive integer k
Source: Dutch BxMO TST 2019 p1
1/10/2020
Prove that for each positive integer
n
n
n
there are at most two pairs
(
a
,
b
)
(a, b)
(
a
,
b
)
of positive integers with following two properties: (i)
a
2
+
b
=
n
a^2 + b = n
a
2
+
b
=
n
, (ii)
a
+
b
a+b
a
+
b
is a power of two, i.e. there is an integer
k
≥
0
k \ge 0
k
≥
0
such that
a
+
b
=
2
k
a+b = 2^k
a
+
b
=
2
k
.
number theory