Find all integer solution(s): 615+x^2 =2^y


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Question Number 61850 by mr W last updated on 10/Jun/19

$F i n d a l l i n t e g e r s o l u t i o n (s) :$ $615 + x^{2} = 2^{y}$
Commented by Rasheed.Sindhi last updated on 10/Jun/19

$O n e s o l u t i o n : x = 59, y = 12$
Commented by Rasheed.Sindhi last updated on 11/Jun/19
$^• 615+x^2 =2^y ⇒2^y ≥615 ⇒2^y ≥1024 ⇒y≥10 ^• 615+x^2 ∈E ⇒x^2 ∈O ⇒x∈O ^• x^2 =2^y −615 ⇒ 2^y −615 is perfect square. 2^(10) −615 not perfect square 2^(11) −615 not perfect square 2^(12) −615=(59)^2 perfect square ∴ x=59,y=12 ^• 2^y −x^2 =615=3×5×41 ⇒ { (( 2^y ≡x^2 (mod 3))),(( 2^y ≡x^2 (mod 5))),(( 2^y ≡x^2 (mod 41))) :} This suggests that 2^y −x^2 has 5 as unit digit and sum of decimal digits is divisible by 3 and it is also divisible by 41. The first point specialy helps ∙....∙ ∙......∙$
$^{∙} 615 + x^{2} = 2^{y}$ $\Rightarrow 2^{y} ⩾ 615$ $\Rightarrow 2^{y} ⩾ 1024$ $\Rightarrow y ⩾ 10$ $^{∙} 615 + x^{2} \in E$ $\Rightarrow x^{2} \in O$ $\Rightarrow x \in O$ $^{∙} x^{2} = 2^{y} - 615$ $\Rightarrow 2^{y} - 615 is perfect square .$ $2^{10} - 615 not perfect square$ $2^{11} - 615 not perfect square$ $2^{12} - 615 = {(59)}^{2} perfect square$ $∴ x = 59, y = 12$ $^{∙} 2^{y} - x^{2} = 615 = 3 \times 5 \times 41$ $\Rightarrow {\begin{cases} 2^{y} \equiv x^{2} (\mod 3) \\ 2^{y} \equiv x^{2} (\mod 5) \\ 2^{y} \equiv x^{2} (\mod 41) \end{cases}$ $This suggests that 2^{y} - x^{2} has$ $5 as unit digit and sum of decimal$ $digits is divisible by 3 and it is also$ $divisible by 41 .$ $T h e f i r s t p o i n t s p e c i a l y h e l p s$ $\cdot . . . . \cdot$ $\cdot . . . . . . \cdot$
Commented by mr W last updated on 10/Jun/19

$t h a n k y o u s i r! n i c e w o r k i n g! c a n w e$ $b e s u r e i f o t h e r s o l u t i o n s e x i s t ?$
Commented by Rasheed.Sindhi last updated on 11/Jun/19

$Sir, {it}^{'} s beyond my capacity .$ $A computer program can help$ $in searching . A program that$ $calculates 2^{y} - 615 for defferent (y, x)$ $systematically and checks for every$ $situation whether {it}^{'} s perfect square$ $or not .$
Commented by Rasheed.Sindhi last updated on 12/Jun/19

$By the way sir, what is the source$ $of this problem ?$
Commented by mr W last updated on 12/Jun/19

$I s a w t h i s q u e s t i o n i n y o u t u b e c h a n n e l$ $‘ ‘ m i n d y o u r {d e c i s i o n}^{″} . b u t i {h a v n}^{'} t$ $w a t c h e d t h e a n s w e r y e t a n d w a n t t o t r y$ $b y m y s e l f a t f i r s t .$
Commented by Rasheed.Sindhi last updated on 12/Jun/19

$U p t o n o w I^{'} v e d i s c o v e r e d o n l y t h i s$ $t h a t y i s e v e n a n d x i s o d d!$ $W e c a n w r i t e t h e e q u a t i o n$ $2^{2 u} - {(2 k + 1)}^{2} = 615$ $a n d t h i s i s t r u e f o r u = 6, k = 29$ $S i r c o u l d t h e g r a p h b e u s e d t o s h o w$ $o t h e r s o l u t i o n s (w h e t h e r t h e y$ $e x i s t o r n o t)$ $∙ ∙ ∙ ∙$ $2^{2 u} - {4 k}^{2} - 4 k - 1 = 615$ $4^{u} - {4 k}^{2} - 4 k = 616$ $4^{u - 1} - k^{2} - k = 154$ $k^{2} + k - 4^{u - 1} = - 154$ $\frac{k (k + 1)}{2} - \frac{2^{2 u - 2}}{2} = - 77$ $\frac{k (k + 1)}{2} - 2^{2 u - 3} = - 77$ $\frac{k (k + 1)}{2} + 77 = 2^{2 u - 3}$ $\frac{k (k + 1)}{2} + 77 \in E$ $\frac{k (k + 1)}{2} \in O$ $4 ∤ k \lor 4 ∤ (k + 1)$ $k is not 4 t or 4 t - 1 type$ $∵ x = 2 k + 1$ $∴ x is not 8 t + 1 or 8 t - 1 (or 8 t + 7) type$ $∵ x \in O$ $∴ x is 8 t + 3 or 8 t - 3 (or 8 t + 5) type .$ $. . . . . .$ $. . . .$
Commented by mr W last updated on 12/Jun/19

$t h a n k y o u s i r .$ $a n y m e t h o d c a n b e u s e d .$ $i^{'} l l t r y t o f o l l o w y o u r s t e p s a n d s e e i f$ $i c a n g o f u r t h e r .$
	Answered by Rasheed.Sindhi last updated on 13/Jun/19
	$^(•1) 615+x^2 =2^y ⇒2^y ≥615⇒2^y ≥1024⇒y≥10 ^(•2) 615+x^2 ∈E ⇒x^2 ∈O⇒x∈O ^(•3) 2^y −615=x^2 ⇒2^y −615 is perfect square. ^(•4) 2^y −x^2 =615=3×5×41 ⇒ { ((2^y ≡x^2 (mod 3))),((2^y ≡x^2 (mod 5))),((2^y ≡x^2 (mod 41))) :} ^(•4.1) 2^y ≡x^2 (mod 5) ⇒2^y (mod 5)=x^2 (mod 5) 2^y (mod 5): { ((^( 2^y ) ),2^0 ,2^1 ,2^2 ,2^3 ,2^4 ,2^5 ,2^6 ,(...)),(^(2^y (md 5)) ,1,2,4,3,1,2,4,(...)) :} 2^(4t) ≡1(mod 5)................(ia) 2^(4t+1) ≡2(mod 5)............(iia) 2^(4t+2) ≡4(mod 5)............(iiia) 2^(4t+3) ≡3 (mod 5)............(iva) (Where t∈W) x^2 (mod 5): (x∈O) { (^( x^2 ) ,1^2 ,3^2 ,5^2 ,7^2 ,9^2 ,(11^2 ),(13^2 ),(15^2 ),(...)),(^(x^2 (md 5)) ,1,4,0,4,1,( 1),( 4),( 0),(...)) :} (10k+1)^2 ≡1(mod 5) (10k+3)^2 ≡4(mod 5) (10k+5)^2 ≡0(mod 5) (10k+7)^2 ≡4(mod 5) (10k+9)^2 ≡1(mod 5) 2^(4t) ≡1(mod 5)→ { (((10k+1)^2 ≡1(mod 5))),(((10k+9)^2 ≡1(mod 5))) :} 2^(4t+2) ≡4(mod 5)→ { (((10k+3)^2 ≡4(mod 5))),(((10k+7)^2 ≡4(mod 5))) :} {: (( 2^(4t+1) ≡2(mod 5))),(( 2^(4t+3) ≡3 (mod 5))) } →No x • y is either 4t or 4t+2 type (y∈E) y→4t type, x→10k+1 or 10k+9 type y→4t+2 type, x→10k+3 or 10k+7 type 2^(4t) −(10k+1)^(2 ) =615 ∨ 2^(4t) −(10k+9)^2 =615; t≥3 ,k≥0 _(x=1,11,21,31.... or 9,19,29,39,49,59,69,...)^(y=12,16,20,...) [ y=12 & x=59 satisfy the equation.] 2^(4t+2) −(10k+3)^(2 ) =615 ∨ 2^(4t+2) −(10k+7)^2 =615; t≥2 ,k≥0 _(x=3,13,23,33,.... or 7,17,27,37,...)^(y=10,14,18,22,....) ...... Continue$
	$^{∙ 1} 615 + x^{2} = 2^{y}$ $\Rightarrow 2^{y} ⩾ 615 \Rightarrow 2^{y} ⩾ 1024 \Rightarrow y ⩾ 10$ $^{∙ 2} 615 + x^{2} \in E$ $\Rightarrow x^{2} \in O \Rightarrow x \in O$ $^{∙ 3} 2^{y} - 615 = x^{2}$ $\Rightarrow 2^{y} - 615 is p e r f e c t s q u a r e .$ $^{∙ 4} 2^{y} - x^{2} = 615 = 3 \times 5 \times 41$ $\Rightarrow {\begin{cases} 2^{y} \equiv x^{2} (\mod 3) \\ 2^{y} \equiv x^{2} (\mod 5) \\ 2^{y} \equiv x^{2} (\mod 41) \end{cases}$ $^{∙ 4.1} 2^{y} \equiv x^{2} (\mod 5)$ $\Rightarrow 2^{y} (\mod 5) = x^{2} (\mod 5)$ $2^{y} (\mod 5) :$ ${\begin{cases} ^{2^{y}} & 2^{0} & 2^{1} & 2^{2} & 2^{3} & 2^{4} & 2^{5} & 2^{6} & . . . \\ ^{2^{y} (md 5)} & 1 & 2 & 4 & 3 & 1 & 2 & 4 & . . . \end{cases}$ $2^{4 t} \equiv 1 (\mod 5) . . . . . . . . . . . . . . . . (ia)$ $2^{4 t + 1} \equiv 2 (\mod 5) . . . . . . . . . . . . (iia)$ $2^{4 t + 2} \equiv 4 (\mod 5) . . . . . . . . . . . . (iiia)$ $2^{4 t + 3} \equiv 3 (\mod 5) . . . . . . . . . . . . (iva)$ $(Where t \in W)$ $x^{2} (\mod 5) : (x \in O)$ ${\begin{cases} ^{x^{2}} & 1^{2} & 3^{2} & 5^{2} & 7^{2} & 9^{2} & 11^{2} & 13^{2} & 15^{2} & . . . \\ ^{x^{2} (md 5)} & 1 & 4 & 0 & 4 & 1 & 1 & 4 & 0 & . . . \end{cases}$ ${(10 k + 1)}^{2} \equiv 1 (\mod 5)$ ${(10 k + 3)}^{2} \equiv 4 (\mod 5)$ ${(10 k + 5)}^{2} \equiv 0 (\mod 5)$ ${(10 k + 7)}^{2} \equiv 4 (\mod 5)$ ${(10 k + 9)}^{2} \equiv 1 (\mod 5)$ $2^{4 t} \equiv 1 (\mod 5) \to {\begin{cases} {(10 k + 1)}^{2} \equiv 1 (\mod 5) \\ {(10 k + 9)}^{2} \equiv 1 (\mod 5) \end{cases}$ $2^{4 t + 2} \equiv 4 (\mod 5) \to {\begin{cases} {(10 k + 3)}^{2} \equiv 4 (\mod 5) \\ {(10 k + 7)}^{2} \equiv 4 (\mod 5) \end{cases}$ $\begin{matrix} 2^{4 t + 1} \equiv 2 (\mod 5) \\ 2^{4 t + 3} \equiv 3 (\mod 5) \end{matrix}} \to No x$ $∙ y is either 4 t or 4 t + 2 type$ $(y \in E)$ $y \to 4 t type, x \to 10 k + 1 or 10 k + 9 type$ $y \to 4 t + 2 type, x \to 10 k + 3 or 10 k + 7 type$ $2^{4 t} - {(10 k + 1)}^{2} = 615$ $\lor 2^{4 t} - {(10 k + 9)}^{2} = 615; t ⩾ 3, k ⩾ 0$ $_{x = 1, 11, 21, 31 . . . . or 9, 19, 29, 39, 49, 59, 69, . . .}^{y = 12, 16, 20, . . .}$ $[y = 12 & x = 59 satisfy the equation .]$ $2^{4 t + 2} - {(10 k + 3)}^{2} = 615$ $\lor 2^{4 t + 2} - {(10 k + 7)}^{2} = 615; t ⩾ 2, k ⩾ 0$ $_{x = 3, 13, 23, 33, . . . . or 7, 17, 27, 37, . . .}^{y = 10, 14, 18, 22, . . . .}$ $. . . . . .$ $Continue$
	Commented by Rasheed.Sindhi last updated on 11/Jun/19
	$⇒ { ((2^y ≡x^2 (mod 3))),((2^y ≡x^2 (mod 5))),((2^y ≡x^2 (mod 41))) :} 2^y ≡x^2 (mod 3) ⇒2^y (mod 3)=x^2 (mod 3) { (( 2^y ),2^0 ,2^1 ,2^2 ,2^3 ),((∽(md 3)),1,2,1,2) :} 2^(2t) ≡1(mod 5) 2^(2t+1) ≡2(mod 5) { (( x^2 ),1^2 ,3^2 ,5^2 ,7^2 ,9^2 ,(11^2 )),((∽(md 3)),1,0,1,1,0,1) :} If y is 2t type, x is 6k−5 or 6k−1 type y=10,12,14,....x=1,7,13,19,..or x=5,11,17,23,.... Continue$
	$\Rightarrow {\begin{cases} 2^{y} \equiv x^{2} (\mod 3) \\ 2^{y} \equiv x^{2} (\mod 5) \\ 2^{y} \equiv x^{2} (\mod 41) \end{cases}$ $2^{y} \equiv x^{2} (\mod 3)$ $\Rightarrow 2^{y} (\mod 3) = x^{2} (\mod 3)$ ${\begin{cases} 2^{y} & 2^{0} & 2^{1} & 2^{2} & 2^{3} \\ ∽ (md 3) & 1 & 2 & 1 & 2 \end{cases}$ $2^{2 t} \equiv 1 (\mod 5)$ $2^{2 t + 1} \equiv 2 (\mod 5)$ ${\begin{cases} x^{2} & 1^{2} & 3^{2} & 5^{2} & 7^{2} & 9^{2} & 11^{2} \\ ∽ (md 3) & 1 & 0 & 1 & 1 & 0 & 1 \end{cases}$ $If y is 2 t type, x is 6 k - 5 or 6 k - 1 type$ $y = 10, 12, 14, . . . . x = 1, 7, 13, 19, . . or$ $x = 5, 11, 17, 23, . . . .$ $Continue$
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