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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
Findallintegersolution(s):615+x2=2y
Commented by Rasheed.Sindhi last updated on 10/Jun/19
One solution:  x=59 , y=12
Onesolution:x=59,y=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           ∙....∙  ∙......∙
615+x2=2y2y6152y1024y10615+x2Ex2OxOx2=2y6152y615isperfectsquare.210615notperfectsquare211615notperfectsquare212615=(59)2perfectsquarex=59,y=122yx2=615=3×5×41{2yx2(mod3)2yx2(mod5)2yx2(mod41)Thissuggeststhat2yx2has5asunitdigitandsumofdecimaldigitsisdivisibleby3anditisalsodivisibleby41.Thefirstpointspecialyhelps.
Commented by mr W last updated on 10/Jun/19
thank you sir! nice working! can we  be sure if other solutions exist?
thankyousir!niceworking!canwebesureifothersolutionsexist?
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.
Sir,itsbeyondmycapacity.Acomputerprogramcanhelpinsearching.Aprogramthatcalculates2y615fordefferent(y,x)systematicallyandchecksforeverysituationwhetheritsperfectsquareornot.
Commented by Rasheed.Sindhi last updated on 12/Jun/19
By the way sir, what is the source  of this problem?
Bythewaysir,whatisthesourceofthisproblem?
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.
Isawthisquestioninyoutubechannelmindyourdecision.butihavntwatchedtheansweryetandwanttotrybymyselfatfirst.
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.               ......        ....
UptonowIvediscoveredonlythisthatyisevenandxisodd!Wecanwritetheequation22u(2k+1)2=615andthisistrueforu=6,k=29Sircouldthegraphbeusedtoshowothersolutions(whethertheyexistornot)22u4k24k1=6154u4k24k=6164u1k2k=154k2+k4u1=154k(k+1)222u22=77k(k+1)222u3=77k(k+1)2+77=22u3k(k+1)2+77Ek(k+1)2O4k4(k+1)kisnot4tor4t1typex=2k+1xisnot8t+1or8t1(or8t+7)typexOxis8t+3or8t3(or8t+5)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.
thankyousir.anymethodcanbeused.illtrytofollowyourstepsandseeificangofurther.
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
1615+x2=2y2y6152y1024y102615+x2Ex2OxO32y615=x22y615isperfectsquare.42yx2=615=3×5×41{2yx2(mod3)2yx2(mod5)2yx2(mod41)4.12yx2(mod5)2y(mod5)=x2(mod5)2y(mod5):{2y202122232425262y(md5)124312424t1(mod5).(ia)24t+12(mod5)(iia)24t+24(mod5)(iiia)24t+33(mod5)(iva)(WheretW)x2(mod5):(xO){x21232527292112132152x2(md5)14041140(10k+1)21(mod5)(10k+3)24(mod5)(10k+5)20(mod5)(10k+7)24(mod5)(10k+9)21(mod5)24t1(mod5){(10k+1)21(mod5)(10k+9)21(mod5)24t+24(mod5){(10k+3)24(mod5)(10k+7)24(mod5)24t+12(mod5)24t+33(mod5)}Noxyiseither4tor4t+2type(yE)y4ttype,x10k+1or10k+9typey4t+2type,x10k+3or10k+7type24t(10k+1)2=61524t(10k+9)2=615;t3,k0x=1,11,21,31.or9,19,29,39,49,59,69,y=12,16,20,[y=12&x=59satisfytheequation.]24t+2(10k+3)2=61524t+2(10k+7)2=615;t2,k0x=3,13,23,33,.or7,17,27,37,y=10,14,18,22,.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
{2yx2(mod3)2yx2(mod5)2yx2(mod41)2yx2(mod3)2y(mod3)=x2(mod3){2y20212223(md3)121222t1(mod5)22t+12(mod5){x21232527292112(md3)101101Ifyis2ttype,xis6k5or6k1typey=10,12,14,.x=1,7,13,19,..orx=5,11,17,23,.Continue

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