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Question-185269

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Question Number 185269 by Rupesh123 last updated on 19/Jan/23
Answered by aleks041103 last updated on 19/Jan/23
p=(√(n−8))+(√(n+8))⇒n≥8 p^2 =2n−2(√(n^2 −64)) ⇒(√(4(n^2 −64)))=2n−p^2 since n,p∈Z⇒(√(4(n^2 −64)))∈Z ⇒∃m∈Z⇒4n^2 −16^2 =m^2 ⇒4∣m^2 ⇒2∣m, so let m→2m ⇒4n^2 −16^2 =4m^2 ⇒n^2 −8^2 =m^2 ⇒(n−m)(n+m)=8^2 =2^6 also n≥8⇒n+m≥8 ⇒n+m=8,16,32,64 n−m=8,4,2,1 so (n,m)∈{(8,0),(10,6),(17,15)} this is necessary but insufficient aftef check n=8⇒p=(√(8−8))+(√(8+8))=4∈Z n=10⇒p=(√(10−8))+(√(10+8))=10(√2)∉Z n=17⇒p=(√(17−8))+(√(17+8))=8∈Z ⇒n∈{8,17} Ans. 25
$${p}=\sqrt{{n}−\mathrm{8}}+\sqrt{{n}+\mathrm{8}}\Rightarrow{n}\geqslant\mathrm{8} \\ $$$${p}^{\mathrm{2}} =\mathrm{2}{n}−\mathrm{2}\sqrt{{n}^{\mathrm{2}} −\mathrm{64}} \\ $$$$\Rightarrow\sqrt{\mathrm{4}\left({n}^{\mathrm{2}} −\mathrm{64}\right)}=\mathrm{2}{n}−{p}^{\mathrm{2}} \\ $$$${since}\:{n},{p}\in\mathbb{Z}\Rightarrow\sqrt{\mathrm{4}\left({n}^{\mathrm{2}} −\mathrm{64}\right)}\in\mathbb{Z} \\ $$$$\Rightarrow\exists{m}\in\mathbb{Z}\Rightarrow\mathrm{4}{n}^{\mathrm{2}} −\mathrm{16}^{\mathrm{2}} ={m}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{4}\mid{m}^{\mathrm{2}} \Rightarrow\mathrm{2}\mid{m},\:{so}\:{let}\:{m}\rightarrow\mathrm{2}{m} \\ $$$$\Rightarrow\mathrm{4}{n}^{\mathrm{2}} −\mathrm{16}^{\mathrm{2}} =\mathrm{4}{m}^{\mathrm{2}} \\ $$$$\Rightarrow{n}^{\mathrm{2}} −\mathrm{8}^{\mathrm{2}} ={m}^{\mathrm{2}} \Rightarrow\left({n}−{m}\right)\left({n}+{m}\right)=\mathrm{8}^{\mathrm{2}} =\mathrm{2}^{\mathrm{6}} \\ $$$${also}\:{n}\geqslant\mathrm{8}\Rightarrow{n}+{m}\geqslant\mathrm{8} \\ $$$$\Rightarrow{n}+{m}=\mathrm{8},\mathrm{16},\mathrm{32},\mathrm{64} \\ $$$${n}−{m}=\mathrm{8},\mathrm{4},\mathrm{2},\mathrm{1} \\ $$$${so} \\ $$$$\left({n},{m}\right)\in\left\{\left(\mathrm{8},\mathrm{0}\right),\left(\mathrm{10},\mathrm{6}\right),\left(\mathrm{17},\mathrm{15}\right)\right\} \\ $$$${this}\:{is}\:{necessary}\:{but}\:{insufficient} \\ $$$${aftef}\:{check} \\ $$$${n}=\mathrm{8}\Rightarrow{p}=\sqrt{\mathrm{8}−\mathrm{8}}+\sqrt{\mathrm{8}+\mathrm{8}}=\mathrm{4}\in\mathbb{Z} \\ $$$${n}=\mathrm{10}\Rightarrow{p}=\sqrt{\mathrm{10}−\mathrm{8}}+\sqrt{\mathrm{10}+\mathrm{8}}=\mathrm{10}\sqrt{\mathrm{2}}\notin\mathbb{Z} \\ $$$${n}=\mathrm{17}\Rightarrow{p}=\sqrt{\mathrm{17}−\mathrm{8}}+\sqrt{\mathrm{17}+\mathrm{8}}=\mathrm{8}\in\mathbb{Z} \\ $$$$\Rightarrow{n}\in\left\{\mathrm{8},\mathrm{17}\right\} \\ $$$${Ans}.\:\mathrm{25} \\ $$$$ \\ $$
Answered by Frix last updated on 19/Jan/23
u=(√(n−8)) ⇒ n=u^2 +8 ⇒ (√(n+8))=(√(u^2 +16))=v∈Z ⇒ u^2 +16=v^2 v^2 −u^2 =16 v=u+w (2u+w)w=16 ⇒ w=1∧2u+1=16 impossible w=2∧2u+2=8 ⇒ u=3 ⇒ n=17 ★ w=4∧2u+4=4 ⇒ u=0 ⇒ n=8 ★ w=8∧2u+8=2 impossible w=16∧2u+16=1 impossible
$${u}=\sqrt{{n}−\mathrm{8}}\:\Rightarrow\:{n}={u}^{\mathrm{2}} +\mathrm{8}\:\Rightarrow \\ $$$$\sqrt{{n}+\mathrm{8}}=\sqrt{{u}^{\mathrm{2}} +\mathrm{16}}={v}\in\mathbb{Z}\:\Rightarrow \\ $$$${u}^{\mathrm{2}} +\mathrm{16}={v}^{\mathrm{2}} \\ $$$${v}^{\mathrm{2}} −{u}^{\mathrm{2}} =\mathrm{16} \\ $$$${v}={u}+{w} \\ $$$$\left(\mathrm{2}{u}+{w}\right){w}=\mathrm{16} \\ $$$$\Rightarrow \\ $$$${w}=\mathrm{1}\wedge\mathrm{2}{u}+\mathrm{1}=\mathrm{16}\:\mathrm{impossible} \\ $$$${w}=\mathrm{2}\wedge\mathrm{2}{u}+\mathrm{2}=\mathrm{8}\:\Rightarrow\:{u}=\mathrm{3}\:\Rightarrow\:{n}=\mathrm{17}\:\bigstar \\ $$$${w}=\mathrm{4}\wedge\mathrm{2}{u}+\mathrm{4}=\mathrm{4}\:\Rightarrow\:{u}=\mathrm{0}\:\:\Rightarrow\:{n}=\mathrm{8}\:\bigstar \\ $$$${w}=\mathrm{8}\wedge\mathrm{2}{u}+\mathrm{8}=\mathrm{2}\:\mathrm{impossible} \\ $$$${w}=\mathrm{16}\wedge\mathrm{2}{u}+\mathrm{16}=\mathrm{1}\:\mathrm{impossible} \\ $$

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