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Problems
Contests
National and Regional Contests
PEN Problems
PEN B Problems
PEN B Problems
Part of
PEN Problems
Subcontests
(6)
7
1
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B 7
Suppose that
p
>
3
p>3
p
>
3
is prime. Prove that the products of the primitive roots of
p
p
p
between
1
1
1
and
p
−
1
p-1
p
−
1
is congruent to
1
1
1
modulo
p
p
p
.
6
1
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B 6
Suppose that
m
m
m
does not have a primitive root. Show that
a
ϕ
(
m
)
2
≡
1
(
m
o
d
m
)
a^{ \frac{\phi(m)}{2}}\equiv 1 \; \pmod{m}
a
2
ϕ
(
m
)
≡
1
(
mod
m
)
for every
a
a
a
relatively prime
m
m
m
.
5
1
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B 5
Let
p
p
p
be an odd prime. If
g
1
,
⋯
,
g
ϕ
(
p
−
1
)
g_{1}, \cdots, g_{\phi(p-1)}
g
1
,
⋯
,
g
ϕ
(
p
−
1
)
are the primitive roots
(
m
o
d
p
)
\pmod{p}
(
mod
p
)
in the range
1
<
g
≤
p
−
1
1<g \le p-1
1
<
g
≤
p
−
1
, prove that
∑
i
=
1
ϕ
(
p
−
1
)
g
i
≡
μ
(
p
−
1
)
(
m
o
d
p
)
.
\sum_{i=1}^{\phi(p-1)}g_{i}\equiv \mu(p-1) \pmod{p}.
i
=
1
∑
ϕ
(
p
−
1
)
g
i
≡
μ
(
p
−
1
)
(
mod
p
)
.
4
1
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B 4
Let
g
g
g
be a Fibonacci primitive root
(
m
o
d
p
)
\pmod{p}
(
mod
p
)
. i.e.
g
g
g
is a primitive root
(
m
o
d
p
)
\pmod{p}
(
mod
p
)
satisfying
g
2
≡
g
+
1
(
m
o
d
p
)
g^2 \equiv g+1\; \pmod{p}
g
2
≡
g
+
1
(
mod
p
)
. Prove that [*]
g
−
1
g-1
g
−
1
is also a primitive root
(
m
o
d
p
)
\pmod{p}
(
mod
p
)
. [*] if
p
=
4
k
+
3
p=4k+3
p
=
4
k
+
3
then
(
g
−
1
)
2
k
+
3
≡
g
−
2
(
m
o
d
p
)
(g-1)^{2k+3} \equiv g-2 \pmod{p}
(
g
−
1
)
2
k
+
3
≡
g
−
2
(
mod
p
)
, and deduce that
g
−
2
g-2
g
−
2
is also a primitive root
(
m
o
d
p
)
\pmod{p}
(
mod
p
)
.
3
1
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B 3
Show that for each odd prime
p
p
p
, there is an integer
g
g
g
such that
1
<
g
<
p
1<g<p
1
<
g
<
p
and
g
g
g
is a primitive root modulo
p
n
p^n
p
n
for every positive integer
n
n
n
.
1
1
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B 1
Let
n
n
n
be a positive integer. Show that there are infinitely many primes
p
p
p
such that the smallest positive primitive root of
p
p
p
is greater than
n
n
n
.