Question

dau 50 de puncte pentru o redactare completa.
p.s. utilizatorii care nu stiu sa nu mi zica asta in raspuns !

Answer 1

$\displaystyle\it\\ecuatia~fiind~simetrica,~putem~considera~p<q<r<s.\\\frac{1}{pqrs} =1-\bigg(\frac{1}{p}+\frac{1}{q}+\frac{1}{r}+\frac{1}{s}\bigg)~\bigg|+\bigg(\frac{1}{p}+\frac{1}{q}+\frac{1}{r}+\frac{1}{s}\bigg).\\1=\frac{1}{pqrs}+\frac{1}{p}+\frac{1}{q}+\frac{1}{r}+\frac{1}{s}<\frac{4}{p}+\frac{1}{p^4} <\frac{5}{p},~de~aici,~evident~p<5.\\daca~p=3 \implies q\geq 5,~r\geq 7~iar~s\geq 11 \implies$

$\displaystyle\it\\\frac{1}{pqrs}+\frac{1}{p}+\frac{1}{q}+\frac{1}{r}+\frac{1}{s}\leq \frac{1}{3\cdot5\cdot7\cdot11} + \frac{1}{3} + \frac{1}{5}+\frac{1}{7}+\frac{1}{11}<\\\frac{1}{3}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}=\frac{1}{3}+\frac{3}{5}<\frac{1}{3}+\frac{2}{3}=1,~asadar~p~nu~poate~fi~3.\\deci,~\boxed{\it p=2}~.\\relatia~din~enunt~devine~:~1=\frac{1}{2qrs}+\frac{1}{2}+\frac{1}{q}+\frac{1}{r}+\frac{1}{s},~\\in~mod~analog~aratam~ca~nu~poate~fi~q\geq 5.\\q\geq 5 \implies r\geq 7~iar~s\geq 11 \implies$

$\displaystyle\it\\\frac{1}{2qrs}+\frac{1}{2}+\frac{1}{q}+\frac{1}{r}+\frac{1}{s} \leq \frac{1}{2\cdot5\cdot7\cdot11} +\frac{1}{2} +\frac{1}{5} +\frac{1}{7}+\frac{1}{11} < 1,\\asadar,~evident~\boxed{\it q=3}~.\\relatia~devine~:~1=\frac{1}{6rs}+\frac{1}{2}+\frac{1}{3}+\frac{1}{r}+\frac{1}{s}=\\1=\frac{1}{6rs}+\frac{5}{6}+\frac{r+s}{rs} \Leftrightarrow 1-\frac{5}{6}=\frac{1}{6rs}+\frac{r+s}{rs} \Leftrightarrow \\\frac{1}{6}=\frac{1+6(r+s)}{6rs},~inmultim~ambii~membri~cu~6.$

$\displaystyle\it\\rs=1+6(r+s) \Leftrightarrow rs-6(r+s)-1=0~\bigg|+37 \implies\\(r-6)(s-6)=37,~rezolvand~usor~ecuatia,~obtinem~\boxed{\it r=7}~si~\boxed{\it s=43}~.\\\boxed{\it (p,q,r,s)\in\left\{(2,3,7,43)\left\},~si~permutarile}~.$