Suppose a, b and n are positive integers, all greater than one. if an+bn is prime, what can you say about n?
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Prime factorization is not needed: we only need the fact that every integer
≠±1
≠±1
has a prime divisor.
Define
r=a/b∈Q.
r=a/b∈Q.
As
b
n
∣
a
n
,
bn∣an,
we know
r
n
∈Z.
rn∈Z.
Write
r=
a
′
/
b
′
r=a′/b′
with
gcd(
a
′
,
b
′
)=1.
gcd(a′,b′)=1.
Let
s=
r
n
.
s=rn.
Then
r
n
=s
rn=s
implies that
(
a
′
)
n
=(
b
′
)
n
⋅s.
(a′)n=(b′)n⋅s.
This shows that every prime divisor of
b
′
b′
divides
(
a
′
)
n
;
(a′)n;
by the definition of a prime, this means that every prime divisor of
b
′
b′
divides
a
′
.
a′.
This contradicts
gcd(
a
′
,
b
′
)=1.
gcd(a′,b′)=1.
Therefore
b
′
b′
has no prime divisor, and is equal to
±1.
±1.
Thus
r=a/b=
a
′
/
b
′
=±
a
′
∈Z.
r=a/b=a′/b′=±a′∈Z.
So
b
b
divides
a.
a.
Hope this helps.
P.S.
The converse is trivial: if
b
b
divides
a,
a,
then
a/b
a/b
is an integer, so
a
n
/
b
n
=(a/b
)
n
an/bn=(a/b)n
is also an integer and
b
n
bn
divides
a
n
.
an.
≠±1
≠±1
has a prime divisor.
Define
r=a/b∈Q.
r=a/b∈Q.
As
b
n
∣
a
n
,
bn∣an,
we know
r
n
∈Z.
rn∈Z.
Write
r=
a
′
/
b
′
r=a′/b′
with
gcd(
a
′
,
b
′
)=1.
gcd(a′,b′)=1.
Let
s=
r
n
.
s=rn.
Then
r
n
=s
rn=s
implies that
(
a
′
)
n
=(
b
′
)
n
⋅s.
(a′)n=(b′)n⋅s.
This shows that every prime divisor of
b
′
b′
divides
(
a
′
)
n
;
(a′)n;
by the definition of a prime, this means that every prime divisor of
b
′
b′
divides
a
′
.
a′.
This contradicts
gcd(
a
′
,
b
′
)=1.
gcd(a′,b′)=1.
Therefore
b
′
b′
has no prime divisor, and is equal to
±1.
±1.
Thus
r=a/b=
a
′
/
b
′
=±
a
′
∈Z.
r=a/b=a′/b′=±a′∈Z.
So
b
b
divides
a.
a.
Hope this helps.
P.S.
The converse is trivial: if
b
b
divides
a,
a,
then
a/b
a/b
is an integer, so
a
n
/
b
n
=(a/b
)
n
an/bn=(a/b)n
is also an integer and
b
n
bn
divides
a
n
.
an.
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