Posted 2 Apr 2013 · 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? 0 Share this post Link to post Share on other sites

0 Posted 2 Apr 2013 · 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 = 36n^{2} + 12n + 12, divisible by 12. (6n - 1)^{2} + 11 = 36n^{2} - 12n + 12, also divisible by 12. Q.E.D. 0 Share this post Link to post Share on other sites

0 Posted 2 Apr 2013 · 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. 0 Share this post Link to post Share on other sites

0 Posted 2 Apr 2013 · 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 = 36n^{2} + 12n + 12, divisible by 12. (6n - 1)^{2} + 11 = 36n^{2} - 12n + 12, also divisible by 12. Q.E.D. 0 Share this post Link to post Share on other sites

0 Posted 2 Apr 2013 (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 = 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 2 Apr 2013 by Prime 0 Share this post Link to post Share on other sites

0 Posted 3 Apr 2013 · 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? 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 0 Share this post Link to post Share on other sites

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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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