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Question Number 61850 by mr W last updated on 10/Jun/19
Find all integer solution(s): 615+x^2 =2^y
$${Find}\:{all}\:{integer}\:{solution}\left({s}\right): \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{615}+\boldsymbol{{x}}^{\mathrm{2}} =\mathrm{2}^{\boldsymbol{{y}}} \\ $$
Commented by Rasheed.Sindhi last updated on 10/Jun/19
One solution: x=59 , y=12
$${One}\:{solution}:\:\:{x}=\mathrm{59}\:,\:{y}=\mathrm{12} \\ $$
Commented by Rasheed.Sindhi last updated on 11/Jun/19
^• 615+x^2 =2^y ⇒2^y ≥615 ⇒2^y ≥1024 ⇒y≥10 ^• 615+x^2 ∈E ⇒x^2 ∈O ⇒x∈O ^• x^2 =2^y −615 ⇒ 2^y −615 is perfect square. 2^(10) −615 not perfect square 2^(11) −615 not perfect square 2^(12) −615=(59)^2 perfect square ∴ x=59,y=12 ^• 2^y −x^2 =615=3×5×41 ⇒ { (( 2^y ≡x^2 (mod 3))),(( 2^y ≡x^2 (mod 5))),(( 2^y ≡x^2 (mod 41))) :} This suggests that 2^y −x^2 has 5 as unit digit and sum of decimal digits is divisible by 3 and it is also divisible by 41. The first point specialy helps ∙....∙ ∙......∙
$$\:^{\bullet} \mathrm{615}+\boldsymbol{{x}}^{\mathrm{2}} =\mathrm{2}^{\boldsymbol{{y}}} \\ $$$$\Rightarrow\mathrm{2}^{{y}} \geqslant\mathrm{615} \\ $$$$\Rightarrow\mathrm{2}^{{y}} \geqslant\mathrm{1024} \\ $$$$\Rightarrow{y}\geqslant\mathrm{10} \\ $$$$\:^{\bullet} \mathrm{615}+\boldsymbol{{x}}^{\mathrm{2}} \in\mathbb{E} \\ $$$$\Rightarrow{x}^{\mathrm{2}} \in\mathbb{O} \\ $$$$\Rightarrow{x}\in\mathbb{O} \\ $$$$\:^{\bullet} {x}^{\mathrm{2}} =\mathrm{2}^{{y}} −\mathrm{615} \\ $$$$\Rightarrow\:\mathrm{2}^{{y}} −\mathrm{615}\:\mathrm{is}\:\mathrm{perfect}\:\mathrm{square}. \\ $$$$\mathrm{2}^{\mathrm{10}} −\mathrm{615}\:\mathrm{not}\:\mathrm{perfect}\:\mathrm{square} \\ $$$$\mathrm{2}^{\mathrm{11}} −\mathrm{615}\:\mathrm{not}\:\mathrm{perfect}\:\mathrm{square} \\ $$$$\mathrm{2}^{\mathrm{12}} −\mathrm{615}=\left(\mathrm{59}\right)^{\mathrm{2}} \:\mathrm{perfect}\:\mathrm{square} \\ $$$$\therefore\:{x}=\mathrm{59},{y}=\mathrm{12} \\ $$$$ \\ $$$$\:^{\bullet} \mathrm{2}^{{y}} −{x}^{\mathrm{2}} =\mathrm{615}=\mathrm{3}×\mathrm{5}×\mathrm{41} \\ $$$$\Rightarrow\begin{cases}{\:\mathrm{2}^{{y}} \equiv{x}^{\mathrm{2}} \left(\mathrm{mod}\:\mathrm{3}\right)}\\{\:\mathrm{2}^{{y}} \equiv{x}^{\mathrm{2}} \left(\mathrm{mod}\:\mathrm{5}\right)}\\{\:\mathrm{2}^{{y}} \equiv{x}^{\mathrm{2}} \left(\mathrm{mod}\:\mathrm{41}\right)}\end{cases} \\ $$$$\mathrm{This}\:\mathrm{suggests}\:\mathrm{that}\:\mathrm{2}^{{y}} −{x}^{\mathrm{2}} \:\mathrm{has} \\ $$$$\mathrm{5}\:\mathrm{as}\:\mathrm{unit}\:\mathrm{digit}\:\mathrm{and}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{decimal} \\ $$$$\mathrm{digits}\:\mathrm{is}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{3}\:\mathrm{and}\:\mathrm{it}\:\mathrm{is}\:\mathrm{also} \\ $$$$\mathrm{divisible}\:\mathrm{by}\:\mathrm{41}. \\ $$$${The}\:{first}\:{point}\:{specialy}\:{helps} \\ $$$$\:\:\: \\ $$$$\:\:\:\:\centerdot….\centerdot \\ $$$$\centerdot……\centerdot \\ $$
Commented by mr W last updated on 10/Jun/19
thank you sir! nice working! can we be sure if other solutions exist?
$${thank}\:{you}\:{sir}!\:{nice}\:{working}!\:{can}\:{we} \\ $$$${be}\:{sure}\:{if}\:{other}\:{solutions}\:{exist}? \\ $$
Commented by Rasheed.Sindhi last updated on 11/Jun/19
Sir, it′s beyond my capacity. A computer program can help in searching.A program that calculates 2^y −615 for defferent (y,x) systematically and checks for every situation whether it′s perfect square or not.
$$\mathrm{Sir},\:\mathrm{it}'\mathrm{s}\:\mathrm{beyond}\:\mathrm{my}\:\mathrm{capacity}. \\ $$$$\mathrm{A}\:\mathrm{computer}\:\mathrm{program}\:\mathrm{can}\:\mathrm{help} \\ $$$$\mathrm{in}\:\mathrm{searching}.\mathrm{A}\:\mathrm{program}\:\mathrm{that} \\ $$$$\mathrm{calculates}\:\mathrm{2}^{{y}} −\mathrm{615}\:\mathrm{for}\:\mathrm{defferent}\:\left({y},{x}\right) \\ $$$$\mathrm{systematically}\:\mathrm{and}\:\mathrm{checks}\:\mathrm{for}\:\mathrm{every} \\ $$$$\mathrm{situation}\:\mathrm{whether}\:\mathrm{it}'\mathrm{s}\:\mathrm{perfect}\:\mathrm{square} \\ $$$$\mathrm{or}\:\mathrm{not}. \\ $$
Commented by Rasheed.Sindhi last updated on 12/Jun/19
By the way sir, what is the source of this problem?
$$\mathrm{By}\:\mathrm{the}\:\mathrm{way}\:\mathrm{sir},\:\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{source} \\ $$$$\mathrm{of}\:\mathrm{this}\:\mathrm{problem}? \\ $$
Commented by mr W last updated on 12/Jun/19
I saw this question in youtube channel “mind your decision”. but i havn′t watched the answer yet and want to try by myself at first.
$${I}\:{saw}\:{this}\:{question}\:{in}\:{youtube}\:{channel} \\ $$$$“{mind}\:{your}\:{decision}''.\:{but}\:{i}\:{havn}'{t} \\ $$$${watched}\:{the}\:{answer}\:{yet}\:{and}\:{want}\:{to}\:{try} \\ $$$${by}\:{myself}\:{at}\:{first}. \\ $$
Commented by Rasheed.Sindhi last updated on 12/Jun/19
Upto now I′ve discovered only this that y is even and x is odd! We can write the equation 2^(2u) −(2k+1)^2 =615 and this is true for u=6, k=29 Sir could the graph be used to show other solutions(whether they exist or not) • • • • 2^(2u) −4k^2 −4k−1=615 4^u −4k^2 −4k=616 4^(u−1) −k^2 −k=154 k^2 +k−4^(u−1) =−154 ((k(k+1))/2)−(2^(2u−2) /2)=−77 ((k(k+1))/2)−2^(2u−3) =−77 ((k(k+1))/2)+77=2^(2u−3) ((k(k+1))/2)+77∈E ((k(k+1))/2)∈O 4∤k ∨ 4∤(k+1) k is not 4t or 4t−1 type ∵x=2k+1 ∴ x is not 8t+1 or 8t−1(or8t+7) type ∵ x∈ O ∴ x is 8t+3 or 8t−3(or 8t+5) type. ...... ....
$${Upto}\:{now}\:{I}'{ve}\:{discovered}\:{only}\:{this} \\ $$$${that}\:{y}\:{is}\:{even}\:{and}\:{x}\:{is}\:{odd}! \\ $$$${We}\:{can}\:{write}\:{the}\:{equation} \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{2}^{\mathrm{2}{u}} −\left(\mathrm{2}{k}+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{615} \\ $$$${and}\:{this}\:{is}\:{true}\:{for}\:{u}=\mathrm{6},\:\mathrm{k}=\mathrm{29} \\ $$$${Sir}\:{could}\:{the}\:{graph}\:{be}\:{used}\:{to}\:\mathrm{s}{how} \\ $$$${other}\:{solutions}\left({whether}\:{they}\right. \\ $$$$\left.{exist}\:{or}\:{not}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\bullet\:\:\bullet\:\:\bullet\:\:\bullet \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}^{\mathrm{2u}} −\mathrm{4k}^{\mathrm{2}} −\mathrm{4k}−\mathrm{1}=\mathrm{615} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{4}^{\mathrm{u}} −\mathrm{4k}^{\mathrm{2}} −\mathrm{4k}=\mathrm{616} \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{4}^{\mathrm{u}−\mathrm{1}} −\mathrm{k}^{\mathrm{2}} −\mathrm{k}=\mathrm{154} \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{k}^{\mathrm{2}} +\mathrm{k}−\mathrm{4}^{\mathrm{u}−\mathrm{1}} =−\mathrm{154} \\ $$$$\:\:\:\:\:\:\:\:\:\frac{\mathrm{k}\left(\mathrm{k}+\mathrm{1}\right)}{\mathrm{2}}−\frac{\mathrm{2}^{\mathrm{2u}−\mathrm{2}} }{\mathrm{2}}=−\mathrm{77} \\ $$$$\:\:\:\:\:\:\:\:\:\frac{\mathrm{k}\left(\mathrm{k}+\mathrm{1}\right)}{\mathrm{2}}−\mathrm{2}^{\mathrm{2u}−\mathrm{3}} =−\mathrm{77} \\ $$$$\:\:\:\:\:\:\:\:\:\frac{\mathrm{k}\left(\mathrm{k}+\mathrm{1}\right)}{\mathrm{2}}+\mathrm{77}=\mathrm{2}^{\mathrm{2u}−\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\frac{\mathrm{k}\left(\mathrm{k}+\mathrm{1}\right)}{\mathrm{2}}+\mathrm{77}\in\mathbb{E} \\ $$$$\:\:\:\:\:\:\:\:\frac{\mathrm{k}\left(\mathrm{k}+\mathrm{1}\right)}{\mathrm{2}}\in\mathbb{O} \\ $$$$\:\:\:\:\:\mathrm{4}\nmid\mathrm{k}\:\vee\:\mathrm{4}\nmid\left(\mathrm{k}+\mathrm{1}\right) \\ $$$$\:\:\:\mathrm{k}\:\mathrm{is}\:\mathrm{not}\:\mathrm{4t}\:\mathrm{or}\:\mathrm{4t}−\mathrm{1}\:\mathrm{type} \\ $$$$\:\:\:\:\:\:\because\mathrm{x}=\mathrm{2k}+\mathrm{1} \\ $$$$\:\:\:\:\:\therefore\:\mathrm{x}\:\mathrm{is}\:\mathrm{not}\:\mathrm{8t}+\mathrm{1}\:\mathrm{or}\:\mathrm{8t}−\mathrm{1}\left(\mathrm{or8t}+\mathrm{7}\right)\:\mathrm{type} \\ $$$$\:\:\:\:\because\:\mathrm{x}\in\:\mathbb{O} \\ $$$$\:\:\:\:\therefore\:\:\mathrm{x}\:\mathrm{is}\:\mathrm{8t}+\mathrm{3}\:\mathrm{or}\:\mathrm{8t}−\mathrm{3}\left(\mathrm{or}\:\mathrm{8t}+\mathrm{5}\right)\:\mathrm{type}. \\ $$$$\:\:\:\:\:\:\: \\ $$$$\:\:\:\:…… \\ $$$$\:\:\:\:\:\:…. \\ $$
Commented by mr W last updated on 12/Jun/19
thank you sir. any method can be used. i′ll try to follow your steps and see if i can go further.
$${thank}\:{you}\:{sir}. \\ $$$${any}\:{method}\:{can}\:{be}\:{used}. \\ $$$${i}'{ll}\:{try}\:{to}\:{follow}\:{your}\:{steps}\:{and}\:{see}\:{if} \\ $$$${i}\:{can}\:{go}\:{further}. \\ $$
Answered by Rasheed.Sindhi last updated on 13/Jun/19
^(•1) 615+x^2 =2^y ⇒2^y ≥615⇒2^y ≥1024⇒y≥10 ^(•2) 615+x^2 ∈E ⇒x^2 ∈O⇒x∈O ^(•3) 2^y −615=x^2 ⇒2^y −615 is perfect square. ^(•4) 2^y −x^2 =615=3×5×41 ⇒ { ((2^y ≡x^2 (mod 3))),((2^y ≡x^2 (mod 5))),((2^y ≡x^2 (mod 41))) :} ^(•4.1) 2^y ≡x^2 (mod 5) ⇒2^y (mod 5)=x^2 (mod 5) 2^y (mod 5): { ((^( 2^y ) ),2^0 ,2^1 ,2^2 ,2^3 ,2^4 ,2^5 ,2^6 ,(...)),(^(2^y (md 5)) ,1,2,4,3,1,2,4,(...)) :} 2^(4t) ≡1(mod 5)................(ia) 2^(4t+1) ≡2(mod 5)............(iia) 2^(4t+2) ≡4(mod 5)............(iiia) 2^(4t+3) ≡3 (mod 5)............(iva) (Where t∈W) x^2 (mod 5): (x∈O) { (^( x^2 ) ,1^2 ,3^2 ,5^2 ,7^2 ,9^2 ,(11^2 ),(13^2 ),(15^2 ),(...)),(^(x^2 (md 5)) ,1,4,0,4,1,( 1),( 4),( 0),(...)) :} (10k+1)^2 ≡1(mod 5) (10k+3)^2 ≡4(mod 5) (10k+5)^2 ≡0(mod 5) (10k+7)^2 ≡4(mod 5) (10k+9)^2 ≡1(mod 5) 2^(4t) ≡1(mod 5)→ { (((10k+1)^2 ≡1(mod 5))),(((10k+9)^2 ≡1(mod 5))) :} 2^(4t+2) ≡4(mod 5)→ { (((10k+3)^2 ≡4(mod 5))),(((10k+7)^2 ≡4(mod 5))) :} {: (( 2^(4t+1) ≡2(mod 5))),(( 2^(4t+3) ≡3 (mod 5))) } →No x • y is either 4t or 4t+2 type (y∈E) y→4t type, x→10k+1 or 10k+9 type y→4t+2 type, x→10k+3 or 10k+7 type 2^(4t) −(10k+1)^(2 ) =615 ∨ 2^(4t) −(10k+9)^2 =615; t≥3 ,k≥0 _(x=1,11,21,31.... or 9,19,29,39,49,59,69,...)^(y=12,16,20,...) [ y=12 & x=59 satisfy the equation.] 2^(4t+2) −(10k+3)^(2 ) =615 ∨ 2^(4t+2) −(10k+7)^2 =615; t≥2 ,k≥0 _(x=3,13,23,33,.... or 7,17,27,37,...)^(y=10,14,18,22,....) ...... Continue
$$\:^{\bullet\mathrm{1}} \:\:\mathrm{615}+\mathrm{x}^{\mathrm{2}} =\mathrm{2}^{\mathrm{y}} \\ $$$$\:\:\:\:\:\:\Rightarrow\mathrm{2}^{\mathrm{y}} \geqslant\mathrm{615}\Rightarrow\mathrm{2}^{\mathrm{y}} \geqslant\mathrm{1024}\Rightarrow\mathrm{y}\geqslant\mathrm{10} \\ $$$$\:^{\bullet\mathrm{2}} \:\mathrm{615}+\mathrm{x}^{\mathrm{2}} \in\mathbb{E} \\ $$$$\:\:\:\:\:\:\Rightarrow\mathrm{x}^{\mathrm{2}} \in\mathbb{O}\Rightarrow\mathrm{x}\in\mathbb{O} \\ $$$$\:^{\bullet\mathrm{3}} \:\:\mathrm{2}^{\mathrm{y}} −\mathrm{615}=\mathrm{x}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\Rightarrow\mathrm{2}^{\mathrm{y}} −\mathrm{615}\:\mathrm{is}\:{perfect}\:{square}. \\ $$$$\:^{\bullet\mathrm{4}} \:\mathrm{2}^{\mathrm{y}} −\mathrm{x}^{\mathrm{2}} =\mathrm{615}=\mathrm{3}×\mathrm{5}×\mathrm{41} \\ $$$$\Rightarrow\begin{cases}{\mathrm{2}^{\mathrm{y}} \equiv\mathrm{x}^{\mathrm{2}} \left(\mathrm{mod}\:\mathrm{3}\right)}\\{\mathrm{2}^{\mathrm{y}} \equiv\mathrm{x}^{\mathrm{2}} \left(\mathrm{mod}\:\mathrm{5}\right)}\\{\mathrm{2}^{\mathrm{y}} \equiv\mathrm{x}^{\mathrm{2}} \left(\mathrm{mod}\:\mathrm{41}\right)}\end{cases} \\ $$$$\:^{\bullet\mathrm{4}.\mathrm{1}} \:\mathrm{2}^{{y}} \equiv{x}^{\mathrm{2}} \left(\mathrm{mod}\:\mathrm{5}\right) \\ $$$$\:\:\:\:\:\Rightarrow\mathrm{2}^{\mathrm{y}} \left(\mathrm{mod}\:\mathrm{5}\right)=\mathrm{x}^{\mathrm{2}} \left(\mathrm{mod}\:\mathrm{5}\right) \\ $$$$\:\:\:\:\:\mathrm{2}^{\mathrm{y}} \left(\mathrm{mod}\:\mathrm{5}\right): \\ $$$$\begin{cases}{\:^{\:\:\:\:\:\:\:\:\mathrm{2}^{\mathrm{y}} } \:}&{\mathrm{2}^{\mathrm{0}} }&{\mathrm{2}^{\mathrm{1}} }&{\mathrm{2}^{\mathrm{2}} }&{\mathrm{2}^{\mathrm{3}} }&{\mathrm{2}^{\mathrm{4}} }&{\mathrm{2}^{\mathrm{5}} }&{\mathrm{2}^{\mathrm{6}} }&{…}\\{\:^{\mathrm{2}^{\mathrm{y}} \left(\mathrm{md}\:\mathrm{5}\right)} }&{\mathrm{1}}&{\mathrm{2}}&{\mathrm{4}}&{\mathrm{3}}&{\mathrm{1}}&{\mathrm{2}}&{\mathrm{4}}&{…}\end{cases} \\ $$$$\:\:\mathrm{2}^{\mathrm{4t}} \equiv\mathrm{1}\left(\mathrm{mod}\:\mathrm{5}\right)…………….\left(\mathrm{ia}\right) \\ $$$$\:\:\mathrm{2}^{\mathrm{4t}+\mathrm{1}} \equiv\mathrm{2}\left(\mathrm{mod}\:\mathrm{5}\right)…………\left(\mathrm{iia}\right) \\ $$$$\:\:\mathrm{2}^{\mathrm{4t}+\mathrm{2}} \equiv\mathrm{4}\left(\mathrm{mod}\:\mathrm{5}\right)…………\left(\mathrm{iiia}\right) \\ $$$$\:\:\mathrm{2}^{\mathrm{4t}+\mathrm{3}} \equiv\mathrm{3}\:\left(\mathrm{mod}\:\mathrm{5}\right)…………\left(\mathrm{iva}\right) \\ $$$$\left(\mathrm{Where}\:\:\mathrm{t}\in\mathbb{W}\right) \\ $$$$\:\:\:\:\:\:\:\mathrm{x}^{\mathrm{2}} \left(\mathrm{mod}\:\mathrm{5}\right):\:\:\:\:\:\left(\mathrm{x}\in\mathbb{O}\right) \\ $$$$\begin{cases}{\:^{\:\:\:\:\:\:\:\mathrm{x}^{\mathrm{2}} } }&{\mathrm{1}^{\mathrm{2}} }&{\mathrm{3}^{\mathrm{2}} }&{\mathrm{5}^{\mathrm{2}} }&{\mathrm{7}^{\mathrm{2}} }&{\mathrm{9}^{\mathrm{2}} }&{\mathrm{11}^{\mathrm{2}} }&{\mathrm{13}^{\mathrm{2}} }&{\mathrm{15}^{\mathrm{2}} }&{…}\\{\:^{\mathrm{x}^{\mathrm{2}} \:\left(\mathrm{md}\:\mathrm{5}\right)} }&{\mathrm{1}}&{\mathrm{4}}&{\mathrm{0}}&{\mathrm{4}}&{\mathrm{1}}&{\:\mathrm{1}}&{\:\:\mathrm{4}}&{\:\:\mathrm{0}}&{…}\end{cases} \\ $$$$\:\:\:\:\left(\mathrm{10k}+\mathrm{1}\right)^{\mathrm{2}} \equiv\mathrm{1}\left(\mathrm{mod}\:\mathrm{5}\right) \\ $$$$\:\:\:\:\left(\mathrm{10k}+\mathrm{3}\right)^{\mathrm{2}} \equiv\mathrm{4}\left(\mathrm{mod}\:\mathrm{5}\right) \\ $$$$\:\:\:\:\left(\mathrm{10k}+\mathrm{5}\right)^{\mathrm{2}} \equiv\mathrm{0}\left(\mathrm{mod}\:\mathrm{5}\right) \\ $$$$\:\:\:\:\left(\mathrm{10k}+\mathrm{7}\right)^{\mathrm{2}} \equiv\mathrm{4}\left(\mathrm{mod}\:\mathrm{5}\right) \\ $$$$\:\:\:\:\left(\mathrm{10k}+\mathrm{9}\right)^{\mathrm{2}} \equiv\mathrm{1}\left(\mathrm{mod}\:\mathrm{5}\right) \\ $$$$ \\ $$$$\:\:\:\mathrm{2}^{\mathrm{4t}} \equiv\mathrm{1}\left(\mathrm{mod}\:\mathrm{5}\right)\rightarrow\begin{cases}{\left(\mathrm{10k}+\mathrm{1}\right)^{\mathrm{2}} \equiv\mathrm{1}\left(\mathrm{mod}\:\mathrm{5}\right)}\\{\left(\mathrm{10k}+\mathrm{9}\right)^{\mathrm{2}} \equiv\mathrm{1}\left(\mathrm{mod}\:\mathrm{5}\right)}\end{cases} \\ $$$$\:\:\:\mathrm{2}^{\mathrm{4t}+\mathrm{2}} \equiv\mathrm{4}\left(\mathrm{mod}\:\mathrm{5}\right)\rightarrow\begin{cases}{\left(\mathrm{10k}+\mathrm{3}\right)^{\mathrm{2}} \equiv\mathrm{4}\left(\mathrm{mod}\:\mathrm{5}\right)}\\{\left(\mathrm{10k}+\mathrm{7}\right)^{\mathrm{2}} \equiv\mathrm{4}\left(\mathrm{mod}\:\mathrm{5}\right)}\end{cases} \\ $$$$\:\:\:\left.\begin{matrix}{\:\mathrm{2}^{\mathrm{4t}+\mathrm{1}} \equiv\mathrm{2}\left(\mathrm{mod}\:\mathrm{5}\right)}\\{\:\mathrm{2}^{\mathrm{4t}+\mathrm{3}} \equiv\mathrm{3}\:\left(\mathrm{mod}\:\mathrm{5}\right)}\end{matrix}\right\}\:\rightarrow\mathrm{No}\:\mathrm{x} \\ $$$$\bullet\:\:\mathrm{y}\:\mathrm{is}\:\mathrm{either}\:\mathrm{4t}\:\mathrm{or}\:\mathrm{4t}+\mathrm{2}\:\mathrm{type} \\ $$$$\:\:\:\left(\mathrm{y}\in\mathbb{E}\right) \\ $$$$\:\:\:\mathrm{y}\rightarrow\mathrm{4t}\:\mathrm{type},\:\mathrm{x}\rightarrow\mathrm{10k}+\mathrm{1}\:\mathrm{or}\:\mathrm{10k}+\mathrm{9}\:\mathrm{type} \\ $$$$\:\:\:\mathrm{y}\rightarrow\mathrm{4t}+\mathrm{2}\:\mathrm{type},\:\mathrm{x}\rightarrow\mathrm{10k}+\mathrm{3}\:\mathrm{or}\:\mathrm{10k}+\mathrm{7}\:\mathrm{type} \\ $$$$\:\:\:\mathrm{2}^{\mathrm{4t}} −\left(\mathrm{10k}+\mathrm{1}\right)^{\mathrm{2}\:\:} =\mathrm{615} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\vee\:\mathrm{2}^{\mathrm{4t}} −\left(\mathrm{10k}+\mathrm{9}\right)^{\mathrm{2}} =\mathrm{615};\:\mathrm{t}\geqslant\mathrm{3}\:,\mathrm{k}\geqslant\mathrm{0} \\ $$$$\:\:_{\mathrm{x}=\mathrm{1},\mathrm{11},\mathrm{21},\mathrm{31}….\:\:\:\:\:\:\:\mathrm{or}\:\:\mathrm{9},\mathrm{19},\mathrm{29},\mathrm{39},\mathrm{49},\mathrm{59},\mathrm{69},…} ^{\mathrm{y}=\mathrm{12},\mathrm{16},\mathrm{20},…} \\ $$$$\:\:\left[\:\mathrm{y}=\mathrm{12}\:\&\:\mathrm{x}=\mathrm{59}\:\mathrm{satisfy}\:\mathrm{the}\:\mathrm{equation}.\right] \\ $$$$\:\:\:\mathrm{2}^{\mathrm{4t}+\mathrm{2}} −\left(\mathrm{10k}+\mathrm{3}\right)^{\mathrm{2}\:\:} =\mathrm{615} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\vee\:\mathrm{2}^{\mathrm{4t}+\mathrm{2}} −\left(\mathrm{10k}+\mathrm{7}\right)^{\mathrm{2}} =\mathrm{615};\:\mathrm{t}\geqslant\mathrm{2}\:,\mathrm{k}\geqslant\mathrm{0} \\ $$$$\:_{\mathrm{x}=\mathrm{3},\mathrm{13},\mathrm{23},\mathrm{33},….\:\:\:\:\mathrm{or}\:\:\mathrm{7},\mathrm{17},\mathrm{27},\mathrm{37},…} ^{\mathrm{y}=\mathrm{10},\mathrm{14},\mathrm{18},\mathrm{22},….} \\ $$$$…… \\ $$$$\:\: \\ $$$$\:\:\:\: \\ $$$$\mathrm{Continue} \\ $$
Commented by Rasheed.Sindhi last updated on 11/Jun/19
⇒ { ((2^y ≡x^2 (mod 3))),((2^y ≡x^2 (mod 5))),((2^y ≡x^2 (mod 41))) :} 2^y ≡x^2 (mod 3) ⇒2^y (mod 3)=x^2 (mod 3) { (( 2^y ),2^0 ,2^1 ,2^2 ,2^3 ),((∽(md 3)),1,2,1,2) :} 2^(2t) ≡1(mod 5) 2^(2t+1) ≡2(mod 5) { (( x^2 ),1^2 ,3^2 ,5^2 ,7^2 ,9^2 ,(11^2 )),((∽(md 3)),1,0,1,1,0,1) :} If y is 2t type, x is 6k−5 or 6k−1 type y=10,12,14,....x=1,7,13,19,..or x=5,11,17,23,.... Continue
$$\Rightarrow\begin{cases}{\mathrm{2}^{\mathrm{y}} \equiv\mathrm{x}^{\mathrm{2}} \left(\mathrm{mod}\:\mathrm{3}\right)}\\{\mathrm{2}^{\mathrm{y}} \equiv\mathrm{x}^{\mathrm{2}} \left(\mathrm{mod}\:\mathrm{5}\right)}\\{\mathrm{2}^{\mathrm{y}} \equiv\mathrm{x}^{\mathrm{2}} \left(\mathrm{mod}\:\mathrm{41}\right)}\end{cases} \\ $$$$\mathrm{2}^{\mathrm{y}} \equiv\mathrm{x}^{\mathrm{2}} \left(\mathrm{mod}\:\mathrm{3}\right) \\ $$$$\Rightarrow\mathrm{2}^{\mathrm{y}} \left(\mathrm{mod}\:\mathrm{3}\right)=\mathrm{x}^{\mathrm{2}} \left(\mathrm{mod}\:\mathrm{3}\right) \\ $$$$\begin{cases}{\:\:\:\:\:\:\mathrm{2}^{\mathrm{y}} }&{\mathrm{2}^{\mathrm{0}} }&{\mathrm{2}^{\mathrm{1}} }&{\mathrm{2}^{\mathrm{2}} }&{\mathrm{2}^{\mathrm{3}} }\\{\backsim\left(\mathrm{md}\:\mathrm{3}\right)}&{\mathrm{1}}&{\mathrm{2}}&{\mathrm{1}}&{\mathrm{2}}\end{cases} \\ $$$$\:\:\:\mathrm{2}^{\mathrm{2t}} \equiv\mathrm{1}\left(\mathrm{mod}\:\mathrm{5}\right) \\ $$$$\:\:\:\mathrm{2}^{\mathrm{2t}+\mathrm{1}} \equiv\mathrm{2}\left(\mathrm{mod}\:\mathrm{5}\right) \\ $$$$\begin{cases}{\:\:\:\:\:\:\:\:\mathrm{x}^{\mathrm{2}} }&{\mathrm{1}^{\mathrm{2}} }&{\mathrm{3}^{\mathrm{2}} }&{\mathrm{5}^{\mathrm{2}} }&{\mathrm{7}^{\mathrm{2}} }&{\mathrm{9}^{\mathrm{2}} }&{\mathrm{11}^{\mathrm{2}} }\\{\backsim\left(\mathrm{md}\:\mathrm{3}\right)}&{\mathrm{1}}&{\mathrm{0}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{0}}&{\mathrm{1}}\end{cases} \\ $$$$\mathrm{If}\:\:\mathrm{y}\:\mathrm{is}\:\mathrm{2t}\:\mathrm{type},\:\mathrm{x}\:\mathrm{is}\:\mathrm{6k}−\mathrm{5}\:\mathrm{or}\:\mathrm{6k}−\mathrm{1}\:\mathrm{type} \\ $$$$\mathrm{y}=\mathrm{10},\mathrm{12},\mathrm{14},….\mathrm{x}=\mathrm{1},\mathrm{7},\mathrm{13},\mathrm{19},..\mathrm{or} \\ $$$$\mathrm{x}=\mathrm{5},\mathrm{11},\mathrm{17},\mathrm{23},…. \\ $$$$\mathrm{Continue} \\ $$

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