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Question Number 3794 by Filup last updated on 21/Dec/15
prove or disprove:     Σ_(i=1) ^n p_1 i=p_2   p_1 ,p_2 ∈P    (1/2)np_1 (n+1)=p_2   (1/2)n(n+1)=(p_2 /p_1 )  n^2 +n=(2/p_1 )p_2     p_i ≠a_1 a_2 ...a_n      a_i ≠1  ∴(2/p_1 )∉Z   p_1 ≠0    n∈Z  n^2 +n−k=0  k=(2/p_1 )p_2   ∴n=((−1±(√(1+4k)))/2)  ∴n=((−1±(√(1+(8/p_1 )p_2 )))/2)  ∴n∉Z    ∴Σ_(i=1) ^n p_1 i≠p_2     Is this correct? Or am I wrong?
$$\mathrm{prove}\:\mathrm{or}\:\mathrm{disprove}:\:\:\:\:\:\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}{p}_{\mathrm{1}} {i}={p}_{\mathrm{2}} \\ $$$${p}_{\mathrm{1}} ,{p}_{\mathrm{2}} \in\mathbb{P} \\ $$$$ \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{np}_{\mathrm{1}} \left({n}+\mathrm{1}\right)={p}_{\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{n}\left({n}+\mathrm{1}\right)=\frac{{p}_{\mathrm{2}} }{{p}_{\mathrm{1}} } \\ $$$${n}^{\mathrm{2}} +{n}=\frac{\mathrm{2}}{{p}_{\mathrm{1}} }{p}_{\mathrm{2}} \\ $$$$ \\ $$$${p}_{{i}} \neq{a}_{\mathrm{1}} {a}_{\mathrm{2}} …{a}_{{n}} \:\:\:\:\:{a}_{{i}} \neq\mathrm{1} \\ $$$$\therefore\frac{\mathrm{2}}{{p}_{\mathrm{1}} }\notin\mathbb{Z}\:\:\:{p}_{\mathrm{1}} \neq\mathrm{0} \\ $$$$ \\ $$$${n}\in\mathbb{Z} \\ $$$${n}^{\mathrm{2}} +{n}−{k}=\mathrm{0} \\ $$$${k}=\frac{\mathrm{2}}{{p}_{\mathrm{1}} }{p}_{\mathrm{2}} \\ $$$$\therefore{n}=\frac{−\mathrm{1}\pm\sqrt{\mathrm{1}+\mathrm{4}{k}}}{\mathrm{2}} \\ $$$$\therefore{n}=\frac{−\mathrm{1}\pm\sqrt{\mathrm{1}+\frac{\mathrm{8}}{{p}_{\mathrm{1}} }{p}_{\mathrm{2}} }}{\mathrm{2}} \\ $$$$\therefore{n}\notin\mathbb{Z} \\ $$$$ \\ $$$$\therefore\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}{p}_{\mathrm{1}} {i}\neq{p}_{\mathrm{2}} \\ $$$$ \\ $$$$\mathrm{Is}\:\mathrm{this}\:\mathrm{correct}?\:\mathrm{Or}\:\mathrm{am}\:\mathrm{I}\:\mathrm{wrong}? \\ $$
Commented by prakash jain last updated on 21/Dec/15
Σ_(i=1) ^n p_1 i=X=((n(n+1)p_1 )/2)  p_1  divides X and  ((n(n+1))/2)  is  an integer(>1 for n>1) divides X.  So X∉P for n>1.
$$\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}{p}_{\mathrm{1}} {i}={X}=\frac{{n}\left({n}+\mathrm{1}\right){p}_{\mathrm{1}} }{\mathrm{2}} \\ $$$${p}_{\mathrm{1}} \:\mathrm{divides}\:\mathrm{X}\:\mathrm{and}\:\:\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}\:\:\mathrm{is} \\ $$$$\mathrm{an}\:\mathrm{integer}\left(>\mathrm{1}\:\mathrm{for}\:{n}>\mathrm{1}\right)\:\mathrm{divides}\:\mathrm{X}. \\ $$$$\mathrm{So}\:\mathrm{X}\notin\mathbb{P}\:\mathrm{for}\:{n}>\mathrm{1}. \\ $$
Commented by prakash jain last updated on 21/Dec/15
Your rational is also correct since if  ((n(n+1))/2)p_1  is a prime than n is not an integer.  so ((n(n+1))/2)p_1 ∉P for ∀n∈N and n>1
$$\mathrm{Your}\:\mathrm{rational}\:\mathrm{is}\:\mathrm{also}\:\mathrm{correct}\:\mathrm{since}\:\mathrm{if} \\ $$$$\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}{p}_{\mathrm{1}} \:\mathrm{is}\:\mathrm{a}\:\mathrm{prime}\:\mathrm{than}\:{n}\:\mathrm{is}\:\mathrm{not}\:\mathrm{an}\:\mathrm{integer}. \\ $$$$\mathrm{so}\:\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}{p}_{\mathrm{1}} \notin\mathbb{P}\:\mathrm{for}\:\forall{n}\in\mathbb{N}\:\mathrm{and}\:{n}>\mathrm{1} \\ $$
Commented by Yozzii last updated on 21/Dec/15
s=Σ_(i=1) ^n i gives triangular numbers and   hence s∈N. Now, p_2 ∈P and p_1 ∈P.  Therefore, sp_1 =p_2  is true iff s=1  and p_1 =p_2 . Otherwise, sp_1 ≠p_2  since  p_2 =p_1 s is suggestive of p_2 ∉P if s>1   or the given equation is an inequation  if p_1 ≠p_2  and s≥1.
$${s}=\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}{i}\:{gives}\:{triangular}\:{numbers}\:{and}\: \\ $$$${hence}\:{s}\in\mathbb{N}.\:{Now},\:{p}_{\mathrm{2}} \in\mathbb{P}\:{and}\:{p}_{\mathrm{1}} \in\mathbb{P}. \\ $$$${Therefore},\:{sp}_{\mathrm{1}} ={p}_{\mathrm{2}} \:{is}\:{true}\:{iff}\:{s}=\mathrm{1} \\ $$$${and}\:{p}_{\mathrm{1}} ={p}_{\mathrm{2}} .\:{Otherwise},\:{sp}_{\mathrm{1}} \neq{p}_{\mathrm{2}} \:{since} \\ $$$${p}_{\mathrm{2}} ={p}_{\mathrm{1}} {s}\:{is}\:{suggestive}\:{of}\:{p}_{\mathrm{2}} \notin\mathbb{P}\:{if}\:{s}>\mathrm{1}\: \\ $$$${or}\:{the}\:{given}\:{equation}\:{is}\:{an}\:{inequation} \\ $$$${if}\:{p}_{\mathrm{1}} \neq{p}_{\mathrm{2}} \:{and}\:{s}\geqslant\mathrm{1}. \\ $$$$ \\ $$
Commented by Yozzii last updated on 21/Dec/15
If p_1 =p_2 ,  n=((−1±(√9))/2)=((±3−1)/2)=1,−2  n∈N⇒n=1 only.  This should be added to the proof  I think.
$${If}\:{p}_{\mathrm{1}} ={p}_{\mathrm{2}} , \\ $$$${n}=\frac{−\mathrm{1}\pm\sqrt{\mathrm{9}}}{\mathrm{2}}=\frac{\pm\mathrm{3}−\mathrm{1}}{\mathrm{2}}=\mathrm{1},−\mathrm{2} \\ $$$${n}\in\mathbb{N}\Rightarrow{n}=\mathrm{1}\:{only}. \\ $$$${This}\:{should}\:{be}\:{added}\:{to}\:{the}\:{proof} \\ $$$${I}\:{think}. \\ $$
Commented by Filup last updated on 21/Dec/15
Thank you!
$$\mathrm{Thank}\:\mathrm{you}! \\ $$

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