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Question-141221

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Question Number 141221 by Engr_Jidda last updated on 16/May/21
Commented by mr W last updated on 16/May/21
no such k exists.
$${no}\:{such}\:{k}\:{exists}. \\ $$
Commented by Engr_Jidda last updated on 16/May/21
how sir??
$${how}\:{sir}?? \\ $$
Answered by MJS_new last updated on 16/May/21
k^2 =x^2 +22 k=x+n∧n≥1 x^2 +2nx+n^2 =x^2 +22 2nx+n^2 =22 x=((22−n^2 )/(2n)) (1) x>0 ⇒ n<5 we only have to try n∈{1, 2, 3, 4} ⇒ ⇒ no integer x exists (2) if you don′t want to try: x∈N ⇒ 2∣(22−n^2 ) ⇒ n=2m x=((22−4m^2 )/(4m))=((11−2m^2 )/(2m)) x∈N ⇒ 2∣(11−2m^2 ) impossible
$${k}^{\mathrm{2}} ={x}^{\mathrm{2}} +\mathrm{22} \\ $$$${k}={x}+{n}\wedge{n}\geqslant\mathrm{1} \\ $$$${x}^{\mathrm{2}} +\mathrm{2}{nx}+{n}^{\mathrm{2}} ={x}^{\mathrm{2}} +\mathrm{22} \\ $$$$\mathrm{2}{nx}+{n}^{\mathrm{2}} =\mathrm{22} \\ $$$${x}=\frac{\mathrm{22}−{n}^{\mathrm{2}} }{\mathrm{2}{n}} \\ $$$$\left(\mathrm{1}\right)\:{x}>\mathrm{0}\:\Rightarrow\:{n}<\mathrm{5}\:\mathrm{we}\:\mathrm{only}\:\mathrm{have}\:\mathrm{to}\:\mathrm{try}\:{n}\in\left\{\mathrm{1},\:\mathrm{2},\:\mathrm{3},\:\mathrm{4}\right\}\:\Rightarrow \\ $$$$\Rightarrow\:\mathrm{no}\:\mathrm{integer}\:{x}\:\mathrm{exists} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{if}\:\mathrm{you}\:\mathrm{don}'\mathrm{t}\:\mathrm{want}\:\mathrm{to}\:\mathrm{try}: \\ $$$${x}\in\mathbb{N}\:\Rightarrow\:\mathrm{2}\mid\left(\mathrm{22}−{n}^{\mathrm{2}} \right)\:\Rightarrow\:{n}=\mathrm{2}{m} \\ $$$${x}=\frac{\mathrm{22}−\mathrm{4}{m}^{\mathrm{2}} }{\mathrm{4}{m}}=\frac{\mathrm{11}−\mathrm{2}{m}^{\mathrm{2}} }{\mathrm{2}{m}} \\ $$$${x}\in\mathbb{N}\:\Rightarrow\:\mathrm{2}\mid\left(\mathrm{11}−\mathrm{2}{m}^{\mathrm{2}} \right)\:\mathrm{impossible} \\ $$
Commented by Engr_Jidda last updated on 20/May/21
thank you sir
$${thank}\:{you}\:{sir} \\ $$
Answered by Rasheed.Sindhi last updated on 17/May/21
k^2 =x^2 +22 k^2 −x^2 =22 (k−x)(k+x)=1×22 ,2×11 { ((k−x=1 ∧ k+x=22)),((k−x=22 ∧ k+x=1)) :}⇒k=11.5∉N { ((k−x=2 ∧ k+x=11)),((k−x=11 ∧ k+x=2)) :}⇒k=6.5∉N No possible k
$${k}^{\mathrm{2}} ={x}^{\mathrm{2}} +\mathrm{22} \\ $$$${k}^{\mathrm{2}} −{x}^{\mathrm{2}} =\mathrm{22} \\ $$$$\left({k}−{x}\right)\left({k}+{x}\right)=\mathrm{1}×\mathrm{22}\:,\mathrm{2}×\mathrm{11} \\ $$$$\begin{cases}{{k}−{x}=\mathrm{1}\:\wedge\:{k}+{x}=\mathrm{22}}\\{{k}−{x}=\mathrm{22}\:\wedge\:{k}+{x}=\mathrm{1}}\end{cases}\Rightarrow{k}=\mathrm{11}.\mathrm{5}\notin\mathbb{N} \\ $$$$\begin{cases}{{k}−{x}=\mathrm{2}\:\wedge\:{k}+{x}=\mathrm{11}}\\{{k}−{x}=\mathrm{11}\:\wedge\:{k}+{x}=\mathrm{2}}\end{cases}\Rightarrow{k}=\mathrm{6}.\mathrm{5}\notin\mathbb{N} \\ $$$${No}\:{possible}\:{k} \\ $$
Commented by Engr_Jidda last updated on 20/May/21
thank you sir
$${thank}\:{you}\:{sir} \\ $$

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