Given that p and q are consecutive odd prime integers, prove that their sum has three or more prime divisors?

Question

not neccesarily distinct

Answer 1

Let P and Q be two consecutive prime integers both greater than 2 with P > Q. Clearly P-Q = 2K for some K.
Now we can write Q = P-2K and the sum of P and Q is then 2P - 2K = 2(P-K) so there will always be a factor of 2. Now, since P and Q are *consecutive* primes, all of the numbers between them (and there are 2K-1 of those numbers) must be composite numbers. Since it is known that P - 2K is prime (it is Q) then it must be that P-K is composite and, from prime factorization, it will have at least 2 prime factors. That makes a total of al least three prime factors as required.


Doug

Answer 2

When you add 2 odd numbers you get an even number. So p + q is divisible by 2. 2 will always be a factor
(p+q)/2 will give you the average of p and q which will be the number in between p and q. Since p and q are odd, the number between them is even. So it is divisible by 2 again.
For p > 3, p is not divisible by 3 (because it is prime) and q is also not divisble by 3. Since p and q are not divisible by 3 the number between p and q must be divisible by 3. Otherwise, you would have 3 numbers in a row not divisible by 3 which is impossible.
Since I excluded p = 3 above you can do that as a special case
let p =3 and q = 5, p+q = 3 + 5 = 8 = 2*2*2

For all other double primes, 2*2*3 is a factor.

Answer 3

consecutive odd numbers are of the form
They are of the form 6n+1, 6n+3
or 6n+3, 6n+5
or 6n-1, 6n+1
(as odd numbers are 6n+1,6n+3,6n+5 same as 6m-1)
1st case 3 has a factor so cannot be a prime unless n =1
so numbers are 1 and 3 so not both prime

case 2
either n = 1 or 1st one is not a prime
if n =1 numbers are 3 and 5 so sum =8 = 2*2*2 (3 odd factors)

case 3
sum = 12n = 2*2*3*n so minimum 3 odd factors

Answer 4

3+5=8
8=2*2*2
= 3 prime integers

Answer 5

Doug, you dont need to add the condition that the prime numbers are greater than 2, if they are odd, they surely are greater than 2. 2 is the only prime number that is even.

Nice demonstration. I liked it

Ana

Answer 6

Doug is correct.

q = p + x, where x is an even number (2y)
q = p + 2y where y is an integer

p+q = p + (p+2y) = 2p+2y = 2(p+y)

p < p+y < q (or x+2y)

since p & q are consecutive prime (no prime in between), (p+y) must not be prime. p+y = ab where a and b are integers

p+q = 2ab