Question Number 4645 by FilupSmith last updated on 18/Feb/16
$${S}=\underset{{i}=\mathrm{1}} {\overset{{x}} {\prod}}\left({i}^{\frac{\mathrm{1}}{\mathrm{2}}\left({x}−{i}+\mathrm{1}\right)\left({x}−{i}+\mathrm{2}\right)} \right),\:\left({x},\:{i}\right)\in\mathbb{Z} \\ $$$$ \\ $$$$\mathrm{Prove}\:\mathrm{that}\:{S}\in\mathbb{Z} \\ $$
Answered by Yozzii last updated on 18/Feb/16
$${Observe}\:{that}\:\mathrm{0}.\mathrm{5}\left({x}−{i}+\mathrm{1}\right)\left({x}−{i}+\mathrm{2}\right)=\underset{{j}=\mathrm{1}} {\overset{{x}−{i}+\mathrm{1}} {\sum}}{j} \\ $$$${which}\:{is}\:{an}\:{integer}\:{where}\:{we}\:{assume}\:{the}\:{index}\:{of}\:{the} \\ $$$${summation}\:{is}\:\mathrm{0}<\mathrm{1}\leqslant{j}\leqslant{x}−{i}+\mathrm{1} \\ $$$${so}\:{x}−{i}+\mathrm{1}>\mathrm{0}\Rightarrow{x}>{i}−\mathrm{1}\geqslant\mathrm{0}\:\Rightarrow{x},{i}\in\mathbb{Z}^{+} . \\ $$$${Therefore},{i}^{\underset{{j}=\mathrm{1}} {\overset{{x}−{i}+\mathrm{1}} {\sum}}{j}} \in\mathbb{Z}^{+} \Rightarrow{S}\in\mathbb{Z}^{+} \:{if}\:{x},{i}\in\mathbb{Z}^{+} . \\ $$$$ \\ $$$$ \\ $$
Commented by FilupSmith last updated on 18/Feb/16
$$\mathrm{A}\:\mathrm{way}\:\mathrm{I}\:\mathrm{thought}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{this}\:\mathrm{was}\:\mathrm{that}: \\ $$$${a}=\left({x}−{i}+\mathrm{1}\right) \\ $$$${b}=\left({x}−{i}+\mathrm{2}\right)\Rightarrow{b}=\left({a}+\mathrm{1}\right) \\ $$$$ \\ $$$$\therefore\:{S}={i}^{\frac{\mathrm{1}}{\mathrm{2}}{ab}} \\ $$$${ab}\:=\:\mathrm{even} \\ $$$$\because\:\mathrm{either}\:{a}\:\mathrm{or}\:{b}\:\mathrm{is}\:\mathrm{even}\:\mathrm{while}\:\mathrm{the}\:\mathrm{other} \\ $$$$\mathrm{is}\:\mathrm{odd}. \\ $$$$ \\ $$$$\therefore\:{ab}=\mathrm{2}{n} \\ $$$$\therefore\:{S}=\sqrt{{i}^{\mathrm{2}{n}} } \\ $$$$\therefore\:{S}\in\mathbb{Z} \\ $$$$ \\ $$$$\therefore\:\underset{{i}=\mathrm{1}} {\overset{{x}−{i}+\mathrm{1}} {\sum}}{i}^{\frac{\mathrm{1}}{\mathrm{2}}\left({x}−{i}+\mathrm{1}\right)\left({x}−{i}+\mathrm{2}\right)} \:\mathrm{is}\:\mathrm{a}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{integers} \\ $$$$\mathrm{in}\:\mathrm{form}: \\ $$$$\underset{{i}=\mathrm{1}} {\overset{{x}−{i}+\mathrm{1}} {\sum}}\:{i}^{{n}} \\ $$$$ \\ $$$$\mathrm{Is}\:\mathrm{this}\:\mathrm{an}\:\mathrm{accurate}\:\mathrm{proof}? \\ $$
Commented by Yozzii last updated on 19/Feb/16
$${The}\:{first}\:{part}\:{isnt}\:{legit},{since}\:{you}\:{left} \\ $$$${out}\:{the}\:\Pi\:{function}\:{in}\:{the}\:{equivalent} \\ $$$${expression}\:{for}\:{S}.\:{The}\:{key}\:{was}\:{your} \\ $$$${realisation}\:{of}\:{i}^{\mathrm{0}.\mathrm{5}{ab}} \:{always}\:{beimg} \\ $$$${an}\:{integer}\:{if}\:{a}\geqslant\mathrm{1},{b}\geqslant\mathrm{1}\:\left({b}={a}+\mathrm{1}\right).\:{So}\:{the}\:{product}\:{is}\:{an}\:{integer}. \\ $$$${Hence}\:{S}\:{is}\:{an}\:{integer}. \\ $$$${If}\:{ab}<\mathrm{0}\:{as}\:{i}\:{varies}\:{you}\:{have}\:{the}\:{i}^{\mathrm{0}.\mathrm{5}{ab}} =\frac{\mathrm{1}}{{i}^{−\mathrm{0}.\mathrm{5}{ab}} } \\ $$$${so}\:{most}\:{terms}\:{in}\:{the}\:{product}\:{will}\:{be} \\ $$$${fractions}.\:{But}\:{I}\:{don}'{t}\:{think}\:{you}\: \\ $$$${meant}\:{for}\:{x}\in\mathbb{Z}^{−} \:{possibly}\:{so}\:{that} \\ $$$$\mathrm{1}\leqslant{i}\leqslant{x}\Rightarrow{ab}>\mathrm{0}. \\ $$$$ \\ $$$$ \\ $$
Commented by FilupSmith last updated on 19/Feb/16
$$\mathrm{I}\:\mathrm{understand},\:\mathrm{thank}\:\mathrm{you}! \\ $$