Section 4.3
87
(a) [BB] We have
a
2
=
p201.
r2
with gcd(p,
r2)
=
1 and
b
=
rf3
s.
Also
gcd(r2,
s)
=
1 since
gcd(r,
s)
=
1.
We conclude that there are two possibilities:
If
f3
=
1, then
gcd(a
2
,
b)
=
p,
while
if
f3
>
1, then
gcd(a
2
,
b)
=
p2.
(b)
Here,
a
3
=
p301.
r3
and
b
=
rf3
s. Now there are three possibilities.
If
f3
=
1, then
gcd(a
3
,
b)
=
p.
If
f3
=
2, then
gcd(a
3
,
b)
=
p2,
and
if
f3
~
3, then
gcd(a
3
,
b)
=
p3.
(c) Here
we
have
a
2
=
p201.
r2
and
b
3
=
p3
f3
s
3.
There are two possibilities.
If
a
=
1, then
gcd(a
2
, b
3
)
=
p2.
If
a>
1, then
gcd(a
2
,b
3
)
=
p3.
29.
[BB]
Let
the prime decompositions
of
a
and
b
be
a
=
pflp~2
...
p~r
and
b
=
qflqg2
...
q~
•.
Since
gcd(a,
b)
=
1, we know that
Pi
=1=
qj
for any
i
and
j.
Thus,
ab
=
pfl
...
p~rqfl
...
q~.
with no
simplification possible. Since
ab
=
x
2
,
it follows that each
ai
and each
f3i
must
be
even. But this
means that
a
and
b
are perfect squares.
30.
(a) We proceed exactly as in the solution to Problem 17. We write
a
and
b
in the form
for certain primes
Pl,
P2,
..
.
,
Pr
and integers
alo
a2,
..
.
,
ar,
f31. f32,
.
..
,
f3r.
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 Summer '10
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 Graph Theory, Prime number, Greatest common divisor, nd

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