Question Number 3794 by Filup last updated on 21/Dec/15
$$\mathrm{prove}\:\mathrm{or}\:\mathrm{disprove}:\:\:\:\:\:\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}{p}_{\mathrm{1}} {i}={p}_{\mathrm{2}} \\ $$$${p}_{\mathrm{1}} ,{p}_{\mathrm{2}} \in\mathbb{P} \\ $$$$ \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{np}_{\mathrm{1}} \left({n}+\mathrm{1}\right)={p}_{\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{n}\left({n}+\mathrm{1}\right)=\frac{{p}_{\mathrm{2}} }{{p}_{\mathrm{1}} } \\ $$$${n}^{\mathrm{2}} +{n}=\frac{\mathrm{2}}{{p}_{\mathrm{1}} }{p}_{\mathrm{2}} \\ $$$$ \\ $$$${p}_{{i}} \neq{a}_{\mathrm{1}} {a}_{\mathrm{2}} ...{a}_{{n}} \:\:\:\:\:{a}_{{i}} \neq\mathrm{1} \\ $$$$\therefore\frac{\mathrm{2}}{{p}_{\mathrm{1}} }\notin\mathbb{Z}\:\:\:{p}_{\mathrm{1}} \neq\mathrm{0} \\ $$$$ \\ $$$${n}\in\mathbb{Z} \\ $$$${n}^{\mathrm{2}} +{n}−{k}=\mathrm{0} \\ $$$${k}=\frac{\mathrm{2}}{{p}_{\mathrm{1}} }{p}_{\mathrm{2}} \\ $$$$\therefore{n}=\frac{−\mathrm{1}\pm\sqrt{\mathrm{1}+\mathrm{4}{k}}}{\mathrm{2}} \\ $$$$\therefore{n}=\frac{−\mathrm{1}\pm\sqrt{\mathrm{1}+\frac{\mathrm{8}}{{p}_{\mathrm{1}} }{p}_{\mathrm{2}} }}{\mathrm{2}} \\ $$$$\therefore{n}\notin\mathbb{Z} \\ $$$$ \\ $$$$\therefore\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}{p}_{\mathrm{1}} {i}\neq{p}_{\mathrm{2}} \\ $$$$ \\ $$$$\mathrm{Is}\:\mathrm{this}\:\mathrm{correct}?\:\mathrm{Or}\:\mathrm{am}\:\mathrm{I}\:\mathrm{wrong}? \\ $$
Commented by prakash jain last updated on 21/Dec/15
$$\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}{p}_{\mathrm{1}} {i}={X}=\frac{{n}\left({n}+\mathrm{1}\right){p}_{\mathrm{1}} }{\mathrm{2}} \\ $$$${p}_{\mathrm{1}} \:\mathrm{divides}\:\mathrm{X}\:\mathrm{and}\:\:\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}\:\:\mathrm{is} \\ $$$$\mathrm{an}\:\mathrm{integer}\left(>\mathrm{1}\:\mathrm{for}\:{n}>\mathrm{1}\right)\:\mathrm{divides}\:\mathrm{X}. \\ $$$$\mathrm{So}\:\mathrm{X}\notin\mathbb{P}\:\mathrm{for}\:{n}>\mathrm{1}. \\ $$
Commented by prakash jain last updated on 21/Dec/15
$$\mathrm{Your}\:\mathrm{rational}\:\mathrm{is}\:\mathrm{also}\:\mathrm{correct}\:\mathrm{since}\:\mathrm{if} \\ $$$$\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}{p}_{\mathrm{1}} \:\mathrm{is}\:\mathrm{a}\:\mathrm{prime}\:\mathrm{than}\:{n}\:\mathrm{is}\:\mathrm{not}\:\mathrm{an}\:\mathrm{integer}. \\ $$$$\mathrm{so}\:\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}{p}_{\mathrm{1}} \notin\mathbb{P}\:\mathrm{for}\:\forall{n}\in\mathbb{N}\:\mathrm{and}\:{n}>\mathrm{1} \\ $$
Commented by Yozzii last updated on 21/Dec/15
$${s}=\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}{i}\:{gives}\:{triangular}\:{numbers}\:{and}\: \\ $$$${hence}\:{s}\in\mathbb{N}.\:{Now},\:{p}_{\mathrm{2}} \in\mathbb{P}\:{and}\:{p}_{\mathrm{1}} \in\mathbb{P}. \\ $$$${Therefore},\:{sp}_{\mathrm{1}} ={p}_{\mathrm{2}} \:{is}\:{true}\:{iff}\:{s}=\mathrm{1} \\ $$$${and}\:{p}_{\mathrm{1}} ={p}_{\mathrm{2}} .\:{Otherwise},\:{sp}_{\mathrm{1}} \neq{p}_{\mathrm{2}} \:{since} \\ $$$${p}_{\mathrm{2}} ={p}_{\mathrm{1}} {s}\:{is}\:{suggestive}\:{of}\:{p}_{\mathrm{2}} \notin\mathbb{P}\:{if}\:{s}>\mathrm{1}\: \\ $$$${or}\:{the}\:{given}\:{equation}\:{is}\:{an}\:{inequation} \\ $$$${if}\:{p}_{\mathrm{1}} \neq{p}_{\mathrm{2}} \:{and}\:{s}\geqslant\mathrm{1}. \\ $$$$ \\ $$
Commented by Yozzii last updated on 21/Dec/15
$${If}\:{p}_{\mathrm{1}} ={p}_{\mathrm{2}} , \\ $$$${n}=\frac{−\mathrm{1}\pm\sqrt{\mathrm{9}}}{\mathrm{2}}=\frac{\pm\mathrm{3}−\mathrm{1}}{\mathrm{2}}=\mathrm{1},−\mathrm{2} \\ $$$${n}\in\mathbb{N}\Rightarrow{n}=\mathrm{1}\:{only}. \\ $$$${This}\:{should}\:{be}\:{added}\:{to}\:{the}\:{proof} \\ $$$${I}\:{think}. \\ $$
Commented by Filup last updated on 21/Dec/15
$$\mathrm{Thank}\:\mathrm{you}! \\ $$