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88
Solutions to Exercises
34. (a) [BB] Write
3n
+
2
=
P1P2
...
Pm
as the product of (necessarily odd) primes. Each of the
Pi
is
either 3 or of the form
3k
+
1 or
3k
+
2. Now we notice that the product of integers of the form
3k
+
1 is of the same form and the product of 3 with such an integer is a multiple of 3. So if none
of the
Pi
is of the form
3k
+
2, the product cannot
be
either.
If
we drop the word odd, the result is
false. For example, 3(2)
+
2
=
8 is divisible by only the prime 2, which is not of the form
3n
+
2
forn
EN.
(b) Write
4n
+
3
=
P1P2··· Pm
as the product of (necessarily odd) primes. Now any odd prime is
of the form
4k
+
1 or
4k
+
3. The product of any two integers of the form
4k
+
1 is of the same
form, so if all the
Pi
were of this form so would be the product, a contradiction.
(c) Write
6n
+
5
=
P1P2
...
Pm
as the product of primes. Each of the
Pi
is of the form
6k
+
1 or
6k
+
5. The product of two integers of the form
6k
+
1 is also of this form, so at least one of the
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This note was uploaded on 11/08/2010 for the course MATH discrete m taught by Professor Any during the Summer '10 term at FSU.
 Summer '10
 any
 Graph Theory, Integers

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