Question Number 178596 by Acem last updated on 18/Oct/22
$${Let}\:\sqrt{{a}}+\:\sqrt{{b}}=\:\sqrt{\mathrm{2023}}\:\:\:,\:{Find}\:{values}\:{of}\:{a},\:{b}\:\in\:\mathbb{N} \\ $$
Commented by mr W last updated on 18/Oct/22
$${a}=\mathrm{7}\left(\mathrm{17}−{n}\right)^{\mathrm{2}} \\ $$$${b}=\mathrm{7}{n}^{\mathrm{2}} \\ $$$${with}\:{n}=\mathrm{1},\mathrm{2},…,\mathrm{16} \\ $$
Commented by Acem last updated on 18/Oct/22
$${I}\:{like}\:{elegant}\:{solutions}! \\ $$$$ \\ $$$${Tmorrow}\:{i}\:{will}\:{try}\:{this}\:{qusetion} \\ $$
Commented by mr W last updated on 19/Oct/22
$${since}\:\mathrm{2023}=\mathrm{7}×\mathrm{17}^{\mathrm{2}} ,\:{so}\:{in}\:{this}\:{case} \\ $$$${the}\:{solution}\:{is}\:{very}\:{obvious}. \\ $$$$\sqrt{{a}}+\sqrt{{b}}=\mathrm{17}\sqrt{\mathrm{7}} \\ $$$${a}=\mathrm{7}{A}^{\mathrm{2}} ,\:{b}=\mathrm{7}{B}^{\mathrm{2}} \\ $$$${A}+{B}=\mathrm{17} \\ $$$${let}\:{B}={n},\:{then}\:{A}=\mathrm{17}−{n} \\ $$$$\Rightarrow{a}=\mathrm{7}\left(\mathrm{17}−{n}\right)^{\mathrm{2}} ,\:{b}=\mathrm{7}{n}^{\mathrm{2}} \\ $$
Commented by mr W last updated on 19/Oct/22
$${and}\:{we}\:{can}\:{also}\:{see} \\ $$$$\sqrt{{a}}+\sqrt{{b}}=\sqrt{\mathrm{2022}}\:{has}\:{no}\:{solution},\:{since} \\ $$$$\mathrm{2022}=\mathrm{2}×\mathrm{3}×\mathrm{337}\:{which}\:{doesn}'{t}\:{contain} \\ $$$${a}\:{perfect}\:{square}. \\ $$
Answered by Rasheed.Sindhi last updated on 19/Oct/22
![{ (((√a) +(√b) =17(√7) ....(i))),(((√a) −(√b) =u(√7) (say)....(ii))) :}[(√a) ,(√b) ≤17(√7) ] (i)+(ii): (√a) =(((17+u)(√7))/2) a=((7(17+u)^2 )/4)≤; u∈O a=((7(17+(2n+1))^2 )/4); n∈Z a=((7(18+2n)^2 )/4); n∈Z determinant ((( a=7(9+n)^2 ≤17(√7) ; n∈Z))) (i)−(ii):(√b) =(((17 −(2n+1))(√7))/2) b=((7(17 −u)^2 )/4); u∈O b=((7(17 −(2n+1))^2 )/4); n∈Z determinant (((b=7(8 −n)^2 ≤17(√7) ; n∈Z))) Continue....](https://www.tinkutara.com/question/Q178620.png)
$$\begin{cases}{\sqrt{{a}}\:+\sqrt{{b}}\:=\mathrm{17}\sqrt{\mathrm{7}}\:….\left({i}\right)}\\{\sqrt{{a}}\:−\sqrt{{b}}\:={u}\sqrt{\mathrm{7}}\:\left({say}\right)….\left({ii}\right)}\end{cases}\left[\sqrt{{a}}\:,\sqrt{{b}}\:\leqslant\mathrm{17}\sqrt{\mathrm{7}}\:\right] \\ $$$$\left({i}\right)+\left({ii}\right):\:\:\sqrt{{a}}\:=\frac{\left(\mathrm{17}+{u}\right)\sqrt{\mathrm{7}}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:{a}=\frac{\mathrm{7}\left(\mathrm{17}+{u}\right)^{\mathrm{2}} }{\mathrm{4}}\leqslant;\:{u}\in\mathbb{O} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:{a}=\frac{\mathrm{7}\left(\mathrm{17}+\left(\mathrm{2}{n}+\mathrm{1}\right)\right)^{\mathrm{2}} }{\mathrm{4}};\:{n}\in\mathbb{Z} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:{a}=\frac{\mathrm{7}\left(\mathrm{18}+\mathrm{2}{n}\right)^{\mathrm{2}} }{\mathrm{4}};\:{n}\in\mathbb{Z} \\ $$$$\begin{array}{|c|}{\:{a}=\mathrm{7}\left(\mathrm{9}+{n}\right)^{\mathrm{2}} \leqslant\mathrm{17}\sqrt{\mathrm{7}}\:\:;\:{n}\in\mathbb{Z}}\\\hline\end{array} \\ $$$$\left({i}\right)−\left({ii}\right):\sqrt{{b}}\:=\frac{\left(\mathrm{17}\:−\left(\mathrm{2}{n}+\mathrm{1}\right)\right)\sqrt{\mathrm{7}}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:{b}=\frac{\mathrm{7}\left(\mathrm{17}\:−{u}\right)^{\mathrm{2}} }{\mathrm{4}};\:{u}\in\mathbb{O} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:{b}=\frac{\mathrm{7}\left(\mathrm{17}\:−\left(\mathrm{2}{n}+\mathrm{1}\right)\right)^{\mathrm{2}} }{\mathrm{4}};\:{n}\in\mathbb{Z} \\ $$$$\begin{array}{|c|}{{b}=\mathrm{7}\left(\mathrm{8}\:−{n}\right)^{\mathrm{2}} \leqslant\mathrm{17}\sqrt{\mathrm{7}}\:;\:{n}\in\mathbb{Z}}\\\hline\end{array}\: \\ $$$${Continue}…. \\ $$