All primes except 3 are congruent to either 1(mod3) or −1(mod3). Watch

isint
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#1
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Hi
I just wanted to reinforce this.

Take 977 then...
977 = 3q' + 1 for some q'
=> 3q' = 976
=> q' = 976/3 = 325.333333333...

but q' is not an integer? How can the above be true?
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Hodor
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(Original post by isint)
Hi
I just wanted to reinforce this.

Take 977 then...
977 = 3q' + 1 for some q'
=> 3q' = 976
=> q' = 976/3 = 325.333333333...

but q' is not an integer? How can the above be true?
You've assumed that 977 is congruent to 1 (mod 3). (It isn't.)
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isint
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But it IS congruent to -1.

977 = 3q' - 1 for some q'
=> 3q' = 978
=> q' = 978/3 = 326 = integer

Is this true for ALL primes?
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Hodor
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(Original post by isint)
But it IS congruent to -1.

977 = 3q' - 1 for some q'
=> 3q' = 978
=> q' = 978/3 = 326 = integer

Is this true for ALL primes?
Every prime greater than 3 is congruent to either 1 or -1 (mod 3), yes. Try to prove it!
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brianeverit
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(Original post by isint)
But it IS congruent to -1.

977 = 3q' - 1 for some q'
=> 3q' = 978
=> q' = 978/3 = 326 = integer

Is this true for ALL primes?
If p is any prime then p-1,p and p+1 are three consecutive integers so one of them must be a multiple of 3 and if p is not equal to 3 then p-1 or p+1 must be divisible by 3 and so must be an integer.
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Boucly
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All numbers are equivalent to 0, 1 or -1 modulo 3. Thinking contrapositive makes this very easy.
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