Question

Does there exist an infinite sequence \(p_0,p_1,p_2,... \) of prime numbers such that
\(p_k\in\{4p_{k-1}-1,\,4p_{k-1}+1\}\qquad (p_k=4p_{k-1}\pm1)\)
for all \(k\in\mathbb{Z}^+ \)

Answer 1

interesting, im thinking if its related to
http://en.wikipedia.org/wiki/Dirichlet's_theorem_on_arithmetic_progressions

Answer 2

may be not as the next term could be 4p+1 or 4p-1 so that kills the arithmetic progression

Answer 3

hmmm

Answer 4

i have a feeling that there does not exist , but still trying to prove

Answer 5

lets see whats going on,

start with \(\large p_0 = 3\) :

\(\large 4*3\pm 1 \) both are primes, pick \(\large p_1 = 11\)

\(\large 4*11- 1 \) is a prime, pick \(\large p_2 = 43\)

\(\large 4*43+1 \) is a prime, pick \(\large p_3 = 173\)

Answer 6

brb

Answer 7

This might be hard to prove or find. Isn't it just a conjecture that there are infinitely many primes of the form 4p+1 .

Answer 8

Clearly it is flip flopping - it has to alternate because, if two consecutive terms are of form 4p+1, then one of them is divisible by 3 :
\[\large 4(4p+1)+1 \equiv p+2 \pmod 3 \]
if \(p \equiv 1 \pmod 3\) then p+2 is divisible by 3 - contradiction
if \(p \equiv -1 \pmod 3\) then ...

Answer 9

not sure 100% of above conclusion, need to tweak a bit..

Answer 10

the previous sequence ends at \(p_5 = 2765\)

http://www.wolframalpha.com/input/?i=is+%284*691%2B1%29+a+prime

Answer 11

so we get a recurrence relation :
\[\large a_{n+1} = 4a_n + (-1)^n\]

solving it gives :
\[\large a_n = a_04^{n-1} + \dfrac{(-1)^n-4^n}{5}\]

Answer 12

the problem simplifies to proving one of these terms is composite

Answer 13

you can conlude like this i hope :
\(\large 4^{n-1} \equiv \pm 1 \pmod 5\) so for some \(n\), the expression \(\large \dfrac{(-1)^n-4^n}{5}\) becomes \(\large \mp 1\mod 5 \) making that term divisible by 5

\(\huge \square \)

Answer 14

Brilliant !

Answer 15

@myininaya 4p+3 or 4p+1 yes its true it gives infinitly many primes , but im talking about a sequense of p's such that blah blah blah

Answer 16

*

Answer 17

looking at about above proof today again freshly -
I see two silly mistakes/blunders

Answer 18

challenge question : figure out the two mistakes :)

Answer 19

haha xD

Answer 20

they can be corrected easily thouhg. .

Answer 21

cant see :'(

Answer 22

guess i should tell about the 2nd mistake :

\[\large \begin{align}a_n &= a_04^{n-1} + \dfrac{(-1)^n-4^n}{5} \\&\equiv \pm a_0 + \dfrac{(-1)^n-4^n}{5} \pmod 5 \end{align} \]

Answer 23

give up xD
ugh tell

Answer 24

that first prime, \(\large a_0\) sticks in there.

Answer 25

so we need to show that above expression gives all the residues in `mod 5` to conclude the proof

Answer 26

simply saying +-1 and -+1 wont do.
need to prove it..

Answer 27

ok then xD

Answer 28

is there an easy way to prove that the expression \(\large\dfrac{(-1)^n-4^n}{5}\) leaves all the residues in mod 5 ?

Answer 29

actually we can forget about \(\large a_0\) because its just a constant..

Answer 30

if the set of elements {x} leave all the residues in mod 5,
{x+c} also leaves all the residues

Answer 31

so proving \(\dfrac{(-1)^n-4^n}{5}\) leaves all the residues in mod 5 is sufficient i think...

Answer 32

hmm im not sure :O

Answer 33

maybe we can work n=0,1,2,.. and see what they evaluate to

Answer 34

\[\large \dfrac{(-1)^1-4^1}{5} \equiv 4\pmod 5\]
\[\large \dfrac{(-1)^2-4^2}{5} \equiv 2\pmod 5\]
\[\large \dfrac{(-1)^6-4^6}{5} \equiv 1\pmod 5\]
etc... im not sure how to prove it without listen them all down at the moment..

Answer 35

hmm well i guess its enough :O

Answer 36

ugh ok il try again when i get free time

Answer 37

mmm ok