BMAD Posted April 2, 2013 Report Share Posted April 2, 2013 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? Quote Link to comment Share on other sites More sharing options...

0 Prime Posted April 2, 2013 Report Share Posted April 2, 2013 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 = 36n^{2} + 12n + 12, divisible by 12. (6n - 1)^{2} + 11 = 36n^{2} - 12n + 12, also divisible by 12. Q.E.D. Quote Link to comment Share on other sites More sharing options...

0 dark_magician_92 Posted April 2, 2013 Report Share Posted April 2, 2013 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. Quote Link to comment Share on other sites More sharing options...

0 BMAD Posted April 2, 2013 Author Report Share Posted April 2, 2013 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 = 36n^{2} + 12n + 12, divisible by 12. (6n - 1)^{2} + 11 = 36n^{2} - 12n + 12, also divisible by 12. Q.E.D. Quote Link to comment Share on other sites More sharing options...

0 Prime Posted April 2, 2013 Report Share Posted April 2, 2013 (edited) 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 = 36n^{2} + 12n + 12, divisible by 12. (6n - 1)^{2} + 11 = 36n^{2} - 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 April 2, 2013 by Prime Quote Link to comment Share on other sites More sharing options...

0 witzar Posted April 3, 2013 Report Share Posted April 3, 2013 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 Quote Link to comment Share on other sites More sharing options...

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## BMAD

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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