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I know that my question might be ambiguous and unclear .

Question : Prove by contradiction that if $p$ is a prime number then it can be written in $p = Sk + 1$ or $p = Sk + 5$ forms .

I don't know what is the meaning of $S$ . If someone explain to me about that , I will appreciate that .

S.H.W
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    Are you sure it is not a poorly printed "$6$"? Sometimes in old books / scanned pdf files it happens. –  May 23 '17 at 20:40
  • If "$S$" is actually $6$, then the fact is true for all prime numbers $p$ except for $2$ and $3$. The djvu format in particular has a way of making crazy substitutions of one character for another, depending on what options were selected in creating the file. – user49640 May 23 '17 at 20:41
  • @G.Sassatelli Unfortunately It is not $6$ and it is $S$ . – S.H.W May 23 '17 at 20:43
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    Where did you get this problem from? Also how do you know it's not $6$? – kingW3 May 23 '17 at 20:43
  • @kingW3 In the my regular and formal book . Now , I asked my friends and they said that they know it is not true but don't have any idea about the right choice . – S.H.W May 23 '17 at 20:48
  • @G.Sassatelli So you are sure that it is $6$ ? Therefore what about $2$ and $3$ which they aren't in those forms ? – S.H.W May 23 '17 at 20:51
  • @S.H.W How could I be sure of anything? I'm not in the head of Prof. Anonymous Mc.Unknow, nor he consulted me when he decided the notation for his Great Book of Mysteries. I just made an occamist claim by observing that there must be something special about the fact that, for some reason, the author does not need stuff like $Sk+3$ or $Sk+7$, and that this almost always occurs $\pmod 6$. –  May 23 '17 at 21:02
  • @G.Sassatelli You think is there a number instead of $S$ that applies the conditions ? – S.H.W May 23 '17 at 21:09
  • You say that it's in your "regular and formal book." I don't understand what that means. kingW3 was asking for the title of the book that the problem came from. – user49640 May 24 '17 at 16:38
  • @user49640 My mean is textbook . As I wrote , the letter $S$ is wrong undoubtedly but I want to know that is there a number instead of $S$ which the question being right ? If we put $6$ , we will lose $2$ and $3$ . Also if you can provide a proof for $S = 6$ and ignore $2$ and $3$ . – S.H.W May 24 '17 at 18:16
  • SHW, if you give the title of the textbook, it's possible someone here will be able to check what the book says. For the proof, every number can be written in one of the forms $6k, 6k + 1, 6k + 2, 6k + 3, 6k + 4, 6k + 5$. Any number in the first, third, fourth or fifth form must be divisible by $2$ or $3$. So it cannot be prime, except if it is $2$ or $3$. – user49640 May 24 '17 at 18:30

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