Question

Prove that if p and p²+8 are prime then so is p³+4.

Answer 1

Every prime p, except 2 and 3 is congruent to 1 or 5 by modulo 6:
p ≡ ±1 (mod 6) implies p² ≡ 1 (mod 6), p² + 8 ≡ 9 ≡ 3 (mod 6), so
p² + 8 is prime only for p = 3. But then p³ + 4 = 31 - also prime.

Answer 2

how u get to know that p =1 or -1 mod 6 ?

Answer 3

bbecause it is always positive

Answer 4

use division algorithm @ikram002p ?

Answer 5

ohhh ic , that make sense

Answer 6

2 | 6k+2
3 | 6k+3
2 | 6k+4

Answer 7

can't be 2, 3, 4, mod six can it?

Answer 8

still , not sure of the proof i mean u can proof its its of the form 6k+1 or 6k+5 but how its possible to make sure its prime ?

Answer 9

its a necessary condition, not a sufficient condition

Answer 10

wait ok ok i got it nw hehehe

Answer 11

Since all primes besides 2 are odd, and since 6k+1, 6k+5 are odd, there will be instances where these two forms generate primes. We're not saying that 6k+1 and 6k+5 are *always* prime.

Answer 12

ok the idea is p^2+8 is only prime when p=3 so we only need to check for p =3

wew ! i freaked out to see proof for prime forfirst look xD

Answer 13

what is modulo thing? I haven't seen that before

Answer 14

well not mudulo , try to think that p^2+8 is only prime when p=3 how could u show that ?

Answer 15

know about division algorithm ?

Answer 16

I'm coming back for this one lol, I haven't put my hand on this for a long time
see ya, please leave this open

Answer 17

use D.A for P=3k+q
since we need prime check for p=3k+1 and p=3k+2
when p=3k+1 then
(3k+1)^2+8=9k^2+6k+9=3(,,,,) not prime
for
when p=3k+2 then
(3k+2)^2+8=9k^2+12k+12=3(,,,,) not prime

so only possible values p=2 or 3 but when its 2 then p^2+8 is not prime then p=3
when p=3 then p^2+8 and p³+4 both primes

Answer 18

@xapproachesinfinity u closed the Qn already :P

Answer 19

nice :) wonder if there is a way to prove this with out using DA

Answer 20

i guess there is hehe , anything leads to composite xD

Answer 21

mitu used the modulo thingy mmm

Answer 22

also since we know P^2+8 divides 3 for all p>3 then that should lead to something ,,,mmm like showing that p^2+8 is a product of 3 consecutive numbers ( don't know how this is gonna work though xD)

Answer 23

yeah DA is good enough i guess :)

Answer 24

i think we can't model a prime number w/o using DA

Answer 25

well mmm

Answer 26

Can we look at any tautologies to help us prove this?

A= p is prime
B= p²+8 is prime
C= p³+4 is prime

So we're saying this right?

\[\Large A \wedge B \implies C\]

This is the same as saying any of the following three I believe?:

\[\Large A \implies (B \implies C) \\ \Large \overline C \implies \overline{(A \wedge B) } \\ \Large \overline C \implies \overline A \vee \overline B \\ \Large \]
I wish I could offer something more helpful than this but I have a feeling that I'm using these incorrectly somehow, since I rarely do proofs.

Answer 27

last two statemetns are easy to see as they rely on demorgan laws

Answer 28

well :D u set up steps to proof xD ( in general )

Answer 29

first statement requires some thought + paper work lol

Answer 30

i think kainui is suggesting to prove it by contrapositive @ikram002p

Answer 31

use truth table to proof first one

Answer 32

yep yep

Answer 33

its like what we did in the proof @ganeshie8 xD
we proved p^2+8 is composite

Answer 34

I'm sure of one thing lol, I only understood the statement that Kainui wrote down

Answer 35

mmm what ur course ? that asked u to prove this ?
or what the subject ?

Answer 36

i was not the one who asked the question, I was just following lol, my course is just precalc
but i had studied all this stuff long ago haha, i'm trying to get things back

Answer 37

ic .. well try to review division algorithm :)
u will got this easily

Answer 38

Yeah I should go and review those stuff to better understand this