Question Number 9049 by Rasheed Soomro last updated on 16/Nov/16
$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{every}\:\mathrm{even}\:\mathrm{number}\:\mathrm{can}\:\mathrm{be}\: \\ $$$$\mathrm{expressed}\:\mathrm{as}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{two}\:\mathrm{primes}\:\mathrm{or} \\ $$$$\mathrm{give}\:\mathrm{an}\:\mathrm{counter}\:\mathrm{example}. \\ $$
Commented by FilupSmith last updated on 16/Nov/16
$$\mathrm{2}{n}={p}_{\mathrm{1}} +{p}_{\mathrm{2}} \\ $$$${p}_{{n}} \in\mathbb{P} \\ $$$$\: \\ $$$$\mathrm{all}\:\mathrm{primes}\:\mathrm{are}\:\mathrm{odd}\:\mathrm{except}\:{p}=\mathrm{2} \\ $$$$\therefore\mathrm{2}{n}=\left(\mathrm{2}{a}+\mathrm{1}\right)+\left(\mathrm{2}{b}+\mathrm{1}\right) \\ $$$$\mathrm{2}{n}=\mathrm{2}{a}+\mathrm{2}{b}+\mathrm{2} \\ $$$$\mathrm{2}{n}=\mathrm{2}\left({a}+{b}+\mathrm{1}\right)\in\mathbb{E} \\ $$
Commented by FilupSmith last updated on 16/Nov/16
$$\mathrm{if}\:{p}=\mathrm{2},\:\mathrm{2}{n}={p}+{p}=\mathrm{2}{p}=\mathrm{4} \\ $$$$\mathrm{2}{n}=\mathrm{4} \\ $$
Commented by Rasheed Soomro last updated on 16/Nov/16
$$\mathrm{All}\:\mathrm{primes}\:\mathrm{are}\:\mathrm{odd},\mathrm{but}\:\mathrm{all}\:\mathrm{odd}\:\mathrm{are} \\ $$$$\mathrm{not}\:\mathrm{necessarily}\:\mathrm{primes}. \\ $$$$\mathrm{So}\:\mathrm{2a}+\mathrm{1}\:\mathrm{and}\:\mathrm{2b}+\mathrm{1}\:\mathrm{here}\:\mathrm{are}\:\mathrm{not} \\ $$$$\mathrm{necessarily}\:\mathrm{prime}. \\ $$
Commented by FilupSmith last updated on 17/Nov/16
$$\mathrm{I}\:\mathrm{dont}\:\mathrm{understand}.\:\mathrm{am}\:\mathrm{i}\:\mathrm{wrong}? \\ $$
Commented by mrW last updated on 23/Nov/16
$$\mathrm{This}\:\mathrm{is}\:\mathrm{the}\:\mathrm{famous}\:\mathrm{Goldbach}'\mathrm{s} \\ $$$$\mathrm{conjecture},\:\mathrm{one}\:\mathrm{of}\:\mathrm{the}\:\mathrm{most} \\ $$$$\mathrm{difficult}\:\mathrm{problems}\:\mathrm{in}\:\mathrm{mathematics}. \\ $$