Prime, prime, prime, prime, prime, prime
What is the smallest positive common difference of a 6 -term arithmetic progression consisting entirely of (positive) prime numbers?
The answer is 30.
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2 solutions
Let the arithmetic progression be as follows:
p
,
p
+
d
,
p
+
2
d
,
p
+
3
d
,
p
+
4
d
,
p
+
5
d
.
We wish to find necessary congruences on
d
. Clearly, the primes 2, 3, and 5 will have no such arithmetic progression. Thus, the possible values of
p
will satisfy:
p
=
1
m
o
d
2
,
p
=
1
,
2
m
o
d
3
,
p
=
1
,
2
,
3
,
4
m
o
d
5
.
However, the only possible congruences for
d
would then be:
d
=
0
m
o
d
2
,
d
=
0
m
o
d
3
,
d
=
0
m
o
d
5
or more simply
d
=
0
m
o
d
3
0
.
As previously stated,
p
>
5
, and thus, we will start with
p
=
7
and
d
=
3
0
.
This gives us the arithmetic progression: 7 , 3 7 , 6 7 , 9 7 , 1 2 7 , 1 5 7 , which are all prime. This must be the smallest such 6-term arithmetic progression.
Why is it clear that 3 and 5 will have no such arithmetic progression? As in, there is no rigorous proof given... Thanks!
If the common difference isn't a multiple of 3, then every third number will be a multiple of 3. If the common difference isn't a multiple of 2 then every other number will be even. If the common difference isn't a multiple of 5 then every fifth number will be a multiple of 5.