Determine-two-distinct-primes-p-and-q-such-that-i-p-q-1-p-q-1-p-q-2-P-All-primes-ii-p-q-1-p-q-1-p-q-2-p-q-2-P-All-primes-

Question Number 17272 by RasheedSoomro last updated on 03/Jul/17

Determine two distinct primes p and q

such that :

(i) p + q + 1, p + q - 1, \frac{p + q}{2} \in P (All primes) ?

(ii) p + q + 1, p + q - 1, \frac{p + q}{2}, \frac{p - q}{2} \in P (All primes) ?

Commented by prakash jain last updated on 03/Jul/17

p + q + 1 is prime .

If p is 6 n + 1, q c a n o t b e 6 m + 1

since p + q + 1 = 6 (n + m) + 3 \notin P

p = 6 n + 1

q = 6 m - 1

p + q - 1 = 6 (n + m) - 1 possibly a prime .

\frac{p + q}{2} is a prime = 3 (n + m) is prime

not possible .

so at least one of p amd q is 3

\Rightarrow q = 3

now we need to find a prime p

4 + p is prime

2 + p is prime

\frac{3 + p}{2} is prime .

\frac{p - 3}{2} is prime .

p = 6 n - 1, p + 4 = 6 n + 3 not a prime .

p = 6 n + 1, p + 2 = 6 n + 3 not a prime .

There are no 2 prime p and q

which satisfy the given four

conditions .

Please review .

Commented by RasheedSoomro last updated on 03/Jul/17

(i) p + q + 1 \Rightarrow 3 + 7 + 1 = 11

(ii) p + q - 1 \Rightarrow 3 + 11 - 1 = 13

(iii) \frac{p + q}{2} \Rightarrow \frac{7 + 31}{2} = 19

(iv) \frac{p - q}{2} \Rightarrow \frac{19 - 13}{2} = 3

More than two conditions to be

satisfied simultaneously is required .

Commented by RasheedSoomro last updated on 04/Jul/17

Lot of thanks for my math - guru!

Commented by RasheedSoomro last updated on 04/Jul/17

I l o v e y o u r s t r a t e g y t o a t t a c k t h e p r o b l e m!

(I had forgotten (6 n + k) - approach, explained

by you in past questions)

N o w I c a n u s e i t f o r t w o / t h r e e c o n d i t i o n s .

I - e \frac{p + q}{2} & \frac{p - q}{2} be both prime etc

Commented by prakash jain last updated on 05/Jul/17

For (i) three conditons

p + q + 1 prime

p + q - 1 prime

\frac{p + q}{2} prime

case p = 6 m + 1, q = 6 n - 1

p + q + 1 not possible

case p = 6 m + 1, q = 6 n + 1

\frac{p + q}{2} not possible

so one of p and q is either 2 or 3 .

2 is ruled out since there is only

one even prime we need sum of

of two distinct primes to be even .

p = 3, q = 6 n + 1

p + q + 1 = 6 n + 4 \notin P

p = 3, q = 6 n - 1

p + q - 1 = 6 n + 2 \notin P

p = q = 2 is only option .