Question

[UNSOLVED] Prove/Disprove : No term in the below infinite sequence is a prime

\(\large 343,~ 34343, ~3434343, ~343434343, \cdots \)

Answer 1

Here is what I have tried so far
prime factorization of first few terms shows some pattern in \(\mod 3\) :

\[
\large \begin{array}\\
\color{Red}{343} &\color{Red}{: 7^3}\\
34343 &: 61\times 563\\
\color{green}{3434343} &\color{green}{: 3\times 11^2\times 9461}\\
\color{red}{343434343} &\color{Red}{: 7\times 521\times 94169}\\
34343434343 &: 47\times 79\times 9249511\\
\color{green}{3434343434343} &\color{green}{: 3^2\times 19\times 29\times 67\times10336531}\\
\color{red}{343434343434343} &\color{Red}{: 7\times 151\times 324914232199}\\
34343434343434343 &: 5638147\times 6091262669
\\
\end{array}
\]

However the \(\large a_{3k+1}\) terms have no apparent pattern yet

Answer 2

you write "43" whatever number of times in front of 343.

Answer 3

Perhaps you should consider doing the same thing you did, but look for patterns among the different adding of just the sequence "434343" instead of just "43" since I imagine there might be a pattern within these starting with 34343.

Is this a programming class? Perhaps you can come across a counter example and just disprove it and save some real effort haha.

Answer 4

I noticed just now that you can always express these numbers as the sum of two number multiplied, that might make it simpler: \[\LARGE 343 = 3(101)+4(10) \\ \LARGE 34343 = 3(10101)+4(1010) \\ \LARGE 3434343 = 3(1010101)+4(101010) \]

Answer 5

thats exactly how it can be concluded that the terms \(\large a_{3k}, a_{3k+2}\) \(k\ge 0\) are composite :

for \(\mod 3\), we consider factors of \(434343\) : 3x7x13x37x43

since \(\large a_{3k+3} = 10^6a_{3k} + 434343\) , if we can show that \(\large a_{3k}, a_{3k+1}, a_{3k+2}\) are divisible by any one of the factors of \(434343\), then we're done. but only two terms have common factors : {3, 7} .

\(\large a_{3k+1}\) terms are diverging forever without any sharing any factors with the \(43434343... \) sequence

Answer 6

the \(\large a_{3k+1}\) terms i am talking about in previous reply are in thick below :

\[\large \begin{array}\\
\color{gray}{343} &\color{gray}{: 7^3}\\
34343 &: 61\times 563\\
\color{gray}{3434343} &\color{gray}{: 3\times 11^2\times 9461}\\
\color{gray}{343434343} &\color{gray}{: 7\times 521\times 94169}\\
34343434343 &: 47\times 79\times 9249511\\
\color{gray}{3434343434343} &\color{gray}{: 3^2\times 19\times 29\times 67\times10336531}\\
\color{gray}{343434343434343} &\color{gray}{: 7\times 151\times 324914232199}\\
34343434343434343 &: 5638147\times 6091262669
\\
\end{array}\]

the terms in gray are divisible by either 3 or 7, so we're done with them.

Answer 7

Yes, that's what I am talking about in my very first response.

Answer 8

yes i get that :) I am not even sure if this is the right approach for handling those \(\large a_{3k+1}\) terms they look very hard >.<

btw, i have tested all the "434343.." sequence upto 30 digits for common factors. No success :(

Answer 9

Ahhh, well I'm trying more stuff. I'm trying to see if I can link the 3k+1 terms to the ones that are multiples of 3 or 7 somehow so that we can get those by some sort of extension. Hmm...

Answer 10

was praying to find counter example :P
but never found , i reached this then it get boring

http://www.wolframalpha.com/input/?i=is+%283434343434343434343434343434343434343434343434343434343434343434343434343434343434343434343434343434343434343434343434343434343434343434343%29+prime+%3F

Answer 11

lol @ikram002p

Answer 12

ikr , this question drive me crazy within these days

Answer 13

At first when I saw this I thought it was just like the same as that other question of yours ikram, showing that 1111, 111111, etc... are all not prime.

Answer 14

:)
(just to keep track of this thread!)

Answer 15

erm about to reveal :P

Answer 16

I want to try to solve this again!

Answer 17

@ganeshie8 xD
we can right then as difference of squares \( x^2-y^2\) s.t \(x-y\neq1\) and x,y are positive natural

Answer 18

them* not then

Answer 19

yes every odd integer can be written as difference of squares,
how is it going to help ?

Answer 20

well , any odd can be written
but its prime when x-y=1 , right ?
then we need to find such relation to show
x-y \neq 1 means its not prime :D

Answer 21

note that we have all of them are 3 mod 4

Answer 22

so we would have
4n+3=(4n+3)^2-(20n+6)
we need some trick to show
for 20n+6 is perfect square for values of n we have hmm

Answer 23

hmm seems mod 4 dont work , i'll try if this gonna work for some mod :|

Answer 24

@Marki (just a digression)
when (x-y)=1, (x-y)(x+y) does not have to be prime, e.g. (5-4)(5+4)=9
but (x-y) ne 1 does imply (x-y)(x+y) is composite. :)

Answer 25

ok ok but the opposite is true
when x-y is not 1 then its not prime , right ?

Answer 26

Absolutely. The above remark does not change any of your subsequent arguments. In fact the contrapositive is when (x-y)(x+y) is prime, (x-y)=1 is probably what you meant.

Answer 27

for 9
try x=3 y=0 instead

Answer 28

yes this is what im standing for :)
showing its not prime not the inverse

Answer 29

wait i made mistake it should be
4n+3=(4n+3)^2-(16n^2+20n+6)

Answer 30

you got \(a_n \equiv 3 \pmod 4\) because the last two digits of every term in the sequence is \(43\) and \(43\equiv 3 \pmod 4\) ?

Answer 31

yes

Answer 32

how did you get
4n+3=(4n+3)^2-(16n^2+20n+6)

?

Answer 33

( 4n+3)^2=16n^2+24 n+9
(4n+3)^2-(4n+3)=16n^2+20n+6

Answer 34

oh just algebra.. fine, im wid u

Answer 35

but im not sure if this work , its just an idea

Answer 36

it wont work because there are infinitely many primes of form 4n+3

so its not possible to factor 4n+3 into two non trivial factors always

Answer 37

no i dont wanna prove in general as i said for n i have only

Answer 38

i would have this sequence as n (note that numbers in black )
8585
8585858585
8585858585858585
8585858585858585858585
......

Answer 39

ohk you're going to relate \(n\) with the terms in our sequence, makes sense..

Answer 40

xD but this one failed from first term im trying to find some other pattern like this hoping :| it works

Answer 41

hmm it dint work for any mod up to 26
i only tried 34343

Answer 42

i think its dumb way to think of any way :|

Answer 43

im not even getting ideas,
you're atleast trying something !

Answer 44

i'll try mod something odd

Answer 45

Another form of representation might help, but it also might not. \[\Large -4 *10^{2n+1}+43 \frac{10^{2n+2}-1}{10^2-1}\]

Answer 46

Playing around and letting x=10^2n I discovered kind of a pattern, but it might not be anything special: \[\Large \frac{1}{99} \left( -4x 10^3 +43x10^2+4x10^1-43* 10^0\right)\] Notice the last term doesn't contain x (or 10 really) but it kind of has this alternating thing going on with the signs and values like maybe it's the solution of some kind of polynomial or something... I don't know.

Answer 47

do u still need help?

Answer 48

\(\)