Question

Challenging Q.
Help! ENCL Question.

Answer 1

this is very challenging!!!

Answer 2

@kropot72 @zzr0ck3r @iambatman ?

Answer 3

aww lol xD
ok the first
non prime number we can got could give us a clue

Answer 4

After that?

Answer 5

wait im thinking :O
ok 2 dont work

Answer 6

Do you have a solid proof, I don't think I am allowed to use trial and error with this. :)

Answer 7

@mathstudent55
@mathmale

Answer 8

what im saying you cant prove if a formula is always prime , so im sure at some point (prime numer ) it must be not prme

Answer 9

@rational

Answer 10

for p prime is odd
p=2m+1
p^6+6p-4=(2m)^6+k(even) + 2m +1+ 12m+6-4 =h+1
mmm no clue lol
i dont know what to say else ;_;

Answer 11

Let say the equation satisfies for some prime \(A\) :
\[p^6 +6p-4 = A\]

Answer 12

\[\implies p^6 +6p-(A+4) = 0\]

Answer 13

Yeah. Then? :D

Answer 14

The only possible integer(rational) roots for this equation are :

\(\pm (A+4)\) and \(\pm \dfrac{A+4}{p}\)

Answer 15

How do we say that @rational ?

Answer 16

rational root theorem gives that ^
let me think if we can use these in a clever way

Answer 17

I think I'd have to look up rational root theorem as well.

Answer 18

http://en.wikipedia.org/wiki/Rational_root_theorem

Answer 19

lol btw, its not *my* theorem :P

Answer 20

so u assume A is prime
mm then A+4/p would be rational right ?

Answer 21

so we need to check
A+4
-(A+4)
??

Answer 22

i think i made a mistake earlier

Answer 23

try to factorise the equation

Answer 24

Not possible. p^6 is too large for any sort of factorization with natural numbers!

Answer 25

could it be all prime number -2 since P is prime not N number ? also im not confinced about this but if we considers that p^6 devide only p ( just thinking)

Answer 26

@KingGeorge

Answer 27

man this question is for someone doing a PHD in maths :)

Answer 28

well @AkashdeepDeb i understand your conserve about using trial and error method , but As i know even the most powerfull formula for prime numbers been disproved using counter exampl so ...

Answer 29

and by try only p=3 give me prime
it shows that
P^6+6P-4 =3k for all p ( and all real number )

Answer 30

so its easy for now to prove that
p^6+6p-4 is composite :)
so
let p =3h+1 or 3h+2 ( which are the possible prime)

Answer 31

wow ! thats brilliant xD

Answer 32

it took me so much time wew ! @rational

Answer 33

I think I do understand!
Do you mind putting it all together?

Answer 34

so case 1 :-
p=3h+1
(3h+1)^6+6(3h+1)-4
evaluate..
i think it should give 3m
case 2 :-
the same but p=3h+2

Answer 35

well lol you do :-\
i give u a hint dont like to give details :P

Answer 36

Yep ! 3h+2 also gives 3n

Answer 37

Why did we choose?

3h
3h+1
3h+2 ?

Answer 38

(3h+2)^6+6(3h+2)-4

3M + 2^6 + 12 -4

3N

Answer 39

thats by division algorithm - 3h, 3h+1, 3h+2 covers all the integers

Answer 40

Yeah, okay.

So do we say that, as p^6+6p-4 is never prime for any Natural number.
There are no primes 'p' for which g(p) = p^6+6p-4 is prime

?

Answer 41

i said P=3 give you prime else dont

Answer 42

Bswan gave u earlier : (3, 743) is the only solution

Answer 43

But '3' falls under '3h' right?
And did we prove that '3h' is not prime?

Answer 44

3h is prime only when h = 1

Answer 45

All other instances are not prime, so they're not of interest

Answer 46

How do we prove that, '3h' is prime only when h=1 and for all other 'h' it is not prime?

Answer 47

here is a short solution :
3h : h=1 is a solution
3h+1 : composite
3h+2 : composite

Answer 48

think a bit, when h > 1
3h = 3(2)

is 3(2) a prime ?

Answer 49

??

I thought we were checking if g(p) = p^6+6p-4 was prime or not!
I see when for 3h, h is even g(p) will be composite.

But what if h = 3 or any other odd number?

Can you tell me what we are actually trying to do here?

Answer 50

you want both p and g(p) to be prime

Answer 51

not just g(p)

Answer 52

p = 3h

for h>1, p is not a prime number

Answer 53

I think I understand! Thank you very much @rational and @BSwan ! :D

Answer 54

@AkashdeepDeb if you proved something for all integer number then that apply for prime as well ...
so its enought to prove its comosite for all real >3

Answer 55

np :)

Answer 56

it was a beautiful idea, thank you very much @BSwan :)