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Brazil Undergrad MO
2022 Brazil Undergrad MO
6
6
Part of
2022 Brazil Undergrad MO
Problems
(1)
Showing tan((p+1) theta) has numerator divisible by p
Source: Brazilian Undergrad Mathematics Olympiad 2022 P6
11/23/2022
Let
p
≡
3
(
mod
4
)
p \equiv 3 \,(\textrm{mod}\, 4)
p
≡
3
(
mod
4
)
be a prime and
θ
\theta
θ
some angle such that
tan
(
θ
)
\tan(\theta)
tan
(
θ
)
is rational. Prove that
tan
(
(
p
+
1
)
θ
)
\tan((p+1)\theta)
tan
((
p
+
1
)
θ
)
is a rational number with numerator divisible by
p
p
p
, that is,
tan
(
(
p
+
1
)
θ
)
=
u
v
\tan((p+1)\theta) = \frac{u}{v}
tan
((
p
+
1
)
θ
)
=
v
u
with
u
,
v
∈
Z
,
v
>
0
,
mdc
(
u
,
v
)
=
1
u, v \in \mathbb{Z}, v >0, \textrm{mdc}(u, v) = 1
u
,
v
∈
Z
,
v
>
0
,
mdc
(
u
,
v
)
=
1
and
u
≡
0
(
mod
p
)
u \equiv 0 \,(\textrm{mod}\,p)
u
≡
0
(
mod
p
)
.
number theory
prime numbers