Number Theory or Algebra? Part 6

Find the sum of all possible positive integer values of p p which makes p , p + 2 , p + 4 p, p+2, p+4 all prime numbers at the same time for each value of p p .


The answer is 3.

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4 solutions

Mursalin Habib
Apr 26, 2014

Well, if p = 3 p=3 then p p , p + 2 p+2 and p + 4 p+4 are all primes. So, p = 3 p=3 works.

Now p p has to be an odd integer, otherwise none of p p , p + 2 p+2 and p + 4 p+4 are primes.

If p p is a positive odd integer greater than 3 3 , then either it is of the form ( 3 k + 1 ) (3k+1) , or it is of the form ( 3 k 1 ) (3k-1) where k k is a positive integer.

If p = 3 k + 1 p=3k+1 , then p + 2 p+2 is divisible by 3 3 .

If p = 3 k 1 p=3k-1 , then p + 4 p+4 is divisible by 3 3 .

So, it is impossible for p p , p + 2 p+2 , p + 4 p+4 to be all primes if p > 3 p>3 .

And our final answer is 3 \boxed{3} [the sum of one integer is the integer itself.

For more information, you could take a look at the first statement of this problem.

thanx ........it was helpful indeed :)

Maitri Mishra - 6 years, 5 months ago

I misundetstand a question., and sum alll p, p+2, p+4 numbers :(

Ela Marinić-Kragić - 5 years, 10 months ago
Masba Islam
May 4, 2014

Python code:

def checkprime(n):

s=0

for i in range(n/2):

    if n%(i+2)==0:

        s=s+1

if s==0:

    return 0

else:

    return 1

def sum(k):

q=0

for p in range(k):

    m=p+2

    a=checkprime(m)+checkprime(m+2)+checkprime(m+4)

    if a==0:

        q=q+m


return q
Akash Deep
Jul 7, 2014

i f w e c o n s i d e r a n y p r i m e p t h e n t h e r e c a n b e 3 c a s e s p 1 ( m o d 3 ) p 2 ( m o d 3 ) o r 3 p 3 w o u l d d i v i d e p o n l y i f p = 3 t r y i n g o u t f o r o t h e r c a s e s : c a s e 1 i f p 1 ( m o d 3 ) t h e n p + 2 3 ( m o d 3 ) w h i c h m e a n s 3 p + 2 b u t i t c a n t b e p o s s i b l e a s p + 2 i s a p r i m e . c a s e 2 i f p 2 ( m o d 3 ) t h e n p + 4 6 ( m o d 3 ) w h i c h m e a n s 3 p + 4 b u t i t c a n t b e p o s s i b l e a s p + 2 i s a p r i m e . f i n a l l y w e f i n d 3 p a n d p i s a p r i m e a s w e l l s o p = 3 if\quad we\quad consider\quad any\quad prime\quad p\quad then\quad there\quad can\quad be\quad 3\quad cases\quad p\quad \equiv \quad 1\quad (mod\quad 3)\\ p\quad \equiv \quad 2\quad (mod\quad 3)\quad or\quad 3|p\quad 3\quad would\quad divide\quad p\quad only\quad if\quad p\quad =\quad 3\\ trying\quad out\quad for\quad other\quad cases\quad :\\ case\quad 1\\ if\quad p\quad \equiv \quad 1\quad (mod\quad 3)\quad then\quad p\quad +\quad 2\quad \equiv \quad 3\quad (mod\quad 3)\quad which\quad means\quad 3\quad |\quad p\quad +\quad 2\\ but\quad it\quad can't\quad be\quad possible\quad as\quad p\quad +\quad 2\quad is\quad a\quad prime.\\ case\quad 2\\ if\quad p\quad \equiv \quad 2\quad (mod\quad 3)\quad then\quad p\quad +\quad 4\quad \equiv \quad 6\quad (mod\quad 3)\quad which\quad means\quad 3\quad |\quad p\quad +\quad 4\\ but\quad it\quad can't\quad be\quad possible\quad as\quad p\quad +\quad 2\quad is\quad a\quad prime.\\ finally\quad we\quad find\quad 3\quad |\quad p\quad and\quad p\quad is\quad a\quad prime\quad as\quad well\quad so\quad p\quad =\quad 3

Basant K Jha
May 2, 2014

except p=3 no other value of p exist which makes p ,p+1 p+2 prime

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