A special case of Dirichlet’s theorem
Jonathan L.F. King
[email protected]
28 October, 2009
(at
15:16
)
Entrance.
An
arithmetic progression
(
A.P.
) means a set
T
+
M
Z
of integers,
♥
1
where the
gap
(
or
modulus
)
M
is a posint
and
translation
(
or
target
)
T
is an integer. I’ll also use
comb
for “arithmetic progression”.
An A.P.
C
:=
T
+
M
Z
is
coprime
if
T
⊥
M
.
1
:
Dirichlet’s Theorem.
Each coprime arithmetic pro
gression contains infinitely many prime numbers.
♦
While this is difficult to prove in general, there are
three easy special
♥
2
cases, the combs
1 + 3
Z
1 + 4
Z
and
1 + 6
Z
.
We will establish this last case.
Henceforth, let
≡
mean
≡
6
and let
“
congruent
”
mean “mod6 congruent”.
2a
:
Lemma.
Suppose a product
q
1
·
q
2
·
. . .
·
q
‘
·
. . .
·
q
L
of
integers
♥
3
is coprime to
6
. Then each multiplicand
q
‘
is coprime to
6
.
♦
Proof.
Exercise
; prove the contrapositive. Where does
your argument use
that each
q
‘
is is an
integer
?
Does the lemma generalize to “6” being replaced
by
N
, an arbitrary posint?
Henceforth, symbols
r
and
q
, with or without ap
pendages, range over the integers.
2b
:
Corollary.
Suppose product
r
:=
q
1
·
q
2
·
. . .
·
q
L
is
congruent to
1
. For oddlymany indices
‘
in
[
1
.. L
]
,
then,
q
‘
≡
1
.
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 Fall '07
 JURY
 Calculus, Integers, Arithmetic progression, Prime number, primes, Jonathan L.F. King

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