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Is this conjecture regarding primes true?

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For every prime number (p) greater than 3, there exists a natural number (n) such that

p^2 = 12(n)-11. Can you provide a counterexample? Else, can you prove it?

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Posted · Report post

All numbers that are prime relative to 2 and 3 are of the form 6n ± 1, where n is an integer.


All prime numbers greater than 3 must be prime relative to 2 and 3 and must be of the form 6n ± 1.
(6n + 1)2 + 11 = 36n2 + 12n + 12, divisible by 12.
(6n - 1)2 + 11 = 36n2 - 12n + 12, also divisible by 12.
Q.E.D.
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Posted · Report post

For every prime number (p) greater than 3, there exists a natural number (n) such that

p^2 = 12(n)-11. Can you provide a counterexample? Else, can you prove it?

All prime nos. >3 can be written as 6k+1 or 6K-1.

So p^2=36k^2+1+12K or -12K. on comparing with 12n-11 or 12(n-1)+1, we get

12(n-1)=12(3k^2 +or-k) hence n-1 = 3k^2+k or 3k^2-k, where k is > or equal to 1, so it shows, its true for all primes.

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Posted · Report post

I never know when to use Q.E.D. vs Without loss of generality, is there a difference or is it just a style thing?

All numbers that are prime relative to 2 and 3 are of the form 6n ± 1, where n is an integer.


All prime numbers greater than 3 must be prime relative to 2 and 3 and must be of the form 6n ± 1.
(6n + 1)2 + 11 = 36n2 + 12n + 12, divisible by 12.
(6n - 1)2 + 11 = 36n2 - 12n + 12, also divisible by 12.
Q.E.D.
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Posted (edited) · Report post

I never know when to use Q.E.D. vs Without loss of generality, is there a difference or is it just a style thing?

All numbers that are prime relative to 2 and 3 are of the form 6n ± 1, where n is an integer.

All prime numbers greater than 3 must be prime relative to 2 and 3 and must be of the form 6n ± 1.

(6n + 1)2 + 11 = 36n2 + 12n + 12, divisible by 12.

(6n - 1)2 + 11 = 36n2 - 12n + 12, also divisible by 12.

Q.E.D.

"Q.E.D" means "which had to be demonstrated" and typically refers to the concluding statement(s) of a proof.

"WLOG" points to a relation between statements/formulas. Like, if it is good for Goose than without loss of generality it must be good for Gender. Or, let's say, in an induction proof: if the rule holds for a randomly chosen number WLOG it must hold for any other number. Q.E.D.

Edited by Prime
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I never know when to use Q.E.D. vs Without loss of generality, is there a difference or is it just a style thing?

WLOG often starts the proof while QED ends it.

You can decide which one to use on the basis of your progress with the proof ;)

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