Question

A great question with a great solution.

Answer 1

What is the sum of all positive prime numbers such that \(n\) such that \(n^2 + 2\) is also a prime?

Answer 2

\[(6k\pm 1)^2 +2 = 3(\text{stuff})\]

so we only need to check primes less than \(6\)

Answer 3

bye

Answer 4

Another long but nice way to work this is to consider using quadratic reciprocity law

Answer 5

where did u get 6k+/- 1

Answer 6

all primes can be expressed in that form

Answer 7

> 3

Answer 8

Every prime is one of the forms 6k+1 or 6k-1

Answer 9

o ok

Answer 10

why

Answer 11

consider all numbers in mod 6

Answer 12

By euclid division algorithm, every integer can be represented in one of the forms

`6k, 6k+1, 6k+2, 6k+3, 6k+4, 6k+5`

Answer 13

yes

Answer 14

eliminate

Answer 15

2 | 6k

2 | 6k+2

3 | 6k+3

2 | 6k+4

Answer 16

okay so 6k+5 same as remainder -1

Answer 17

hmm

Answer 18

what would this do for this

Answer 19

so we have established that a prime can only be of form 6k+1 or 6k-1

Answer 20

oh nvm

Answer 21

ok i got it now

Answer 22

i see so there are other forms like this from different mods

Answer 23

But why mod 6?

Answer 24

it's a well-known result. the \(\pm 1\) is a great tool.

Answer 25

we need that 1^2 out

Answer 26

so we know its a factor of 3

Answer 27

wonder if there is a proof to show this only happens in mod 6

Answer 28

I don't quite understand that question.
"such that n such that n^2+2 is also a prime" ???

Answer 29

actually ya that is kinda broken english lol

Answer 30

Such that both n and n^2+2 are primes?

Answer 31

yes

Answer 32

Sorry, I meant this:

What is the sum of all primes \(n\) such that \(n^2 + 2\) is also a prime?

Answer 33

still doesnt make sense fully

Answer 34

are u saying n is a sum of primes?

Answer 35

all the primes?

Answer 36

We need to find the sum of all primes n such that both n and n^2+2 are primes.

Answer 37

can i jut see where u got the question actually

Answer 38

p1^2+2 + p2^2+2 + p3^2+2...
so we are trying to see the prime numbers where this sum is still prime right

Answer 39

and all the numbers greater than 6 we say are already factors of 3

Answer 40

saw*

Answer 41

You just want to find the sum of the prime numbers that satisfy this statement.

Answer 42

I just got it on a facebook page.

Answer 43

Another fun problem

find the sum of all integers \(n\) such that \(n^4+4\) is prime

Answer 44

1 works.
even numbers don't.

Answer 45

consecutive integers of any random integers?

Answer 46

im thinking we need to see one that has +/-2 in mod now

Answer 47

mod 5...
cases

5k: gotta check this one
5k + 1: can't be prime
5k + 2: can't be prime
5k + 3: can't be prime
5k + 4: can't be prime

Answer 48

that looks neat ! XD
there is another shortcut for this too

Answer 49

\[n^4+4 = (n^2)^2 + 4 = (n^2)^2 + 4n^2 + 4 - 4n^2 = (n^2+2)^2 - 4n^2 = \cdots \]

Answer 50

\[(5k)^4 + 4 = 625k^4 + 4 = 625k^4 + 100k^2 + 4 - 100k^2 \]\[= (25k + 4)^2 - (10k)^2 = \cdots\]

Answer 51

holy lol why didn't i think of factoring like this in the general case

Answer 52

5k+2 can be prime

Answer 53

so can 5k+3 and 5k+1 and 5k+4

Answer 54

the reason mod 6 works so well is that 6 is the product of the lowest 2 primes

Answer 55

\[\large (5k+a)^4 + 4\equiv a^4 + 4 \pmod{5} \]

and Fermat's little them gives us

\[\large a^4\equiv 1 \pmod{5}\]

Answer 56

Fermat's little thm requires \(5\nmid a\)

so we only need to check the case \(a = 5k\)

Answer 57

@Mimi_x3 what i meant was \(n^4 + 4 \) can't be prime for 5k + 2.

Answer 58

i'm so dumb. why didn't i think of factoring in the general case

Answer 59

ah okay

Answer 60

nice little fermat application