Find-all-integer-solution-s-615-x-2-2-y-

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 + \boldsymbol x^{2} = 2^{\boldsymbol 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 + \boldsymbol x^{2} = 2^{\boldsymbol y}

\Rightarrow 2^{y} ⩾ 615

\Rightarrow 2^{y} ⩾ 1024

\Rightarrow y ⩾ 10

^{∙} 615 + \boldsymbol 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 \dots . \cdot

\cdot \dots \dots \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 .

\dots \dots

\dots .

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}

\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} & \dots \\ ^{2^{y} (md 5)} & 1 & 2 & 4 & 3 & 1 & 2 & 4 & \dots \end{cases}

2^{4 t} \equiv 1 (\mod 5) \dots \dots \dots \dots \dots . (ia)

2^{4 t + 1} \equiv 2 (\mod 5) \dots \dots \dots \dots (iia)

2^{4 t + 2} \equiv 4 (\mod 5) \dots \dots \dots \dots (iiia)

2^{4 t + 3} \equiv 3 (\mod 5) \dots \dots \dots \dots (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} & \dots \\ ^{x^{2} (md 5)} & 1 & 4 & 0 & 4 & 1 & 1 & 4 & 0 & \dots \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 \dots . or 9, 19, 29, 39, 49, 59, 69, \dots}^{y = 12, 16, 20, \dots}

[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, \dots . or 7, 17, 27, 37, \dots}^{y = 10, 14, 18, 22, \dots .}

\dots \dots

Continue

Commented by Rasheed.Sindhi last updated on 11/Jun/19

\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, \dots . x = 1, 7, 13, 19, . . or

x = 5, 11, 17, 23, \dots .

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