Question

The number of integers n > 1, such that n, n + 2, n + 4 are all prime numbers, is

Answer 1

Heya User,

We claim :-> For every prime integer 'n' > 3,
-> 'n' is of the form --> 6k ± 1 ...

Let 'n' > 3 be 6k ± 1

--> n + 2 = 6k + 3  || 6k + 1 
--> n + 4 = 6k + 5  || 6k + 3

=> We see -->> 6k + 3 = 3[ 2k + 1 ] which obviously has more than one factors except when for [ k = 0 ]

So, either way, n = 6k ± 1 won't let the rest numbers be primes 0_0
                      --> [ One bad fish can spoil all :p ]

However, if k = 0, then n = 6k ± 1 = ± 1 which is a contradiction to our assumption [ n > 3 ]

==> n ≤ 3, where at n = 3 is the only value that satisfies the above question..

--> n = 3 || n + 2 = 5 || n + 4 = 7
--> [ n , n + 2 , n + 4 ] = [ 3 , 5 , 7 ]

Hence, there exists only one prime '3' for which n , n + 2 , n + 4 are all primes ^_^