Primal Progression
How many positive integers 0 < p < 1 0 0 0 exist such that the arithmetic progression starting with p and with a difference 1 0 0 has more than two prime numbers as consecutive terms in it.
The answer is 0.
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3 solutions
Let us assume that p , p + 1 0 0 , p + 2 0 0 are the consecutive primes that we are talking about.Now,we know that every prime is of the form 6 x ± 1 (except for 2 and 3 ,you can test these by hand).Let us assume that p = 6 x + 1 then, p + 2 0 0 becomes 6 x + 2 0 1 which has 3 as a factor.Now assume that p = 6 x − 1 then, p + 1 0 0 = 6 x + 9 9 which has 3 as a factor hence the naswer is 0 .Cheers!
Simply if the primes are p, p +100 and p + 200. Then by congruence if p is congruent to k mod 3. Then p+100 and P+200 are congruent to k+1 and k+2 mod 3. So one of them is divisible by 3. and prime divisible by 3 is only 3. So 3, 103 and 203 can be a solution but 203 is composite. Hence nosolution i.e. 0
L e t p b e t h e f i r s t p r i m e . S o 3 ∤ p . S o 3 ∣ ( p + 1 o r p + 2 ) B u t 1 0 0 ≡ 1 ( m o d 3 ) , a n d 2 0 0 ≡ 2 ( m o d 3 ) . S o p + 1 0 0 ≡ p + 1 ( m o d 3 ) , a n d p + 2 0 0 ≡ 2 ( m o d 3 ) . ⟹ 3 ∣ ( p + 1 0 0 o r p + 2 0 0 ) . S o o n e o f p + 1 0 0 , o r p + 2 0 0 i s n o t a p r i m e . ⟹ t h r e e c o n s e c u t i v e p r i m e s i n A P w i t h a d i f f e r e n c e o f 1 0 0 n o t p o s s i b l e .
More than two prime numbers means, atleast 3
Let, The number P be prime xxxx
+100 next number is x(x+1)xx
+100 next number is x(x+2)xx
Sum of the digits of these 3 numbers is 4x,4x+1,4x+2
For no value of X the three are divisible by 3.
One number is definitely divisible by 3. So, no 3 consecutive are prime.
Why do you assume that all the digits of the first number are equal? Secondly, let P be say 9999, then P+100 = 10099 and P+200 = 10199. The sum of digits are 36, 19, 20 . So, the sum of digits logic is flawed even if P has repeating digits.
Consider that there are three prime numbers q , q + 1 0 0 , q + 2 0 0 appearing in the arithmetic progression starting with some p .
It is obvious that q is an odd number and q = 1 (as 1 is not a prime number).
Let q = x m o d 3 , then ( q + 1 0 0 ) = ( x + 9 9 + 1 ) m o d 3 = ( x + 1 ) m o d 3
Similarly, ( q + 2 0 0 ) = ( x + 2 ) m o d 3 .
If q > 3 , then one of the numbers x , x + 1 , x + 2 would be 0 modulo 3 and hence composite. Thus, our assumption that there are three consecutive prime numbers is wrong.
If q = 3 then ( q , q + 1 0 0 , q + 2 0 0 ) = ( 3 , 1 0 3 , 2 0 3 ) . But, here too 2 0 3 = 7 × 2 9 . Thus, there is no artithmetic progression starting from a non negative integer, and having a difference 100, that can have three prime numbers as consecutive terms.