Matematické Fórum

stuart clark · 09. 08. 2011 04:36

determin all pair $(x,y)$ of integer such that $1+2^x+2^{2x+1}=y^2$

pietro · 12. 08. 2011 20:12

VaK · 12. 08. 2011 22:16

Dobrý den,
$y$ must be odd : $y=2z+1$ . After adjustment $2^{x-2}(2^{x+1}+1)=z(z+1)$
then
1) $z=2^{x-2}p , z+1=\frac{2^{x+1}+1}{p}$ , p is integer
$2^{x-2}p=\frac{2^{x+1}+1-p}{p}$
$2^{x-2}p^2+p-(8*2^{x-2}+1)=0$
for $2^{x-2}p >> 1$ is approximatly
$2^{x-2}p^2 = 8*2^{x-2} => p=\sqrt{8} =>p=3$
let $2^{x-2}=a$
$9a+3-8a-1=0 , a=-2$ => for this case is not a solution.
2) $z+1=2^{x-2}p , z=\frac{2^{x+1}+1}{p}$ , p is integer
$2^{x-2}p^2-p-(8*2^{x-2}+1)=0$
as in case 1) is here p=3
$9a-3-8a-1=0 , a=4 , x=4 , z=11 , y=23$
It follows that x=0 , y=2 and x=4 , y=23 are the only
solution of the equation.

stuart clark · 14. 08. 2011 06:23

Autor skryl část textu. Zobrazit/skrýt!

check_drummer · 14. 08. 2011 22:02

VaK napsal(a):
for $2^{x-2}p >> 1$ is approximatly ...

How large should x and p be for using them in the approximation?
(Because as I understand it - this approximation can be only used for large enough x and p, am I right?)

VaK · 15. 08. 2011 19:51

Dobrý den,
from
$2^{x-2}p^2-p-(8*2^{x-2}+1)=0$ (in case 2))
is $p=\frac{1\pm \sqrt{1+4a(8a+1)}}{2a}$
then
x a p ap p^2
3 2 3.176 6.352 10.088
4 4 3.000 12.000 9.000
5 8 2.913 23.308 8.489
6 16 2.871 45.934 8.242
...............................
inf. inf. sqrt(8) inf. 8.000
2.828
it follows that p is integer only for x=4.

Matematické Fórum

#1 09. 08. 2011 04:36

oair of integer

#2 12. 08. 2011 20:12

Re: oair of integer

#3 12. 08. 2011 22:16

Re: oair of integer

#4 14. 08. 2011 06:23

Re: oair of integer

#5 14. 08. 2011 22:02

Re: oair of integer

VaK napsal(a):

#6 15. 08. 2011 19:51

Re: oair of integer

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