Solving the Equation $2(2^a-1)a^2+2=2^{a+1}-(2^{a^2}-2)a$

[anemone]

Solve the equation $2(2^a-1)a^2+2=2^{a+1}-(2^{a^2}-2)a$ for real numbers $a$.

 

[Albert1]

Solve the equation $2(2^a-1)a^2+2=2^{a+1}-(2^{a^2}-2)a$ for real numbers $a$.

Spoiler

it is clear that $a=0$ is a solution
let :$a^2=b, 2^a=c,$ we have :
$2bc-2b+2=2c-2^ba+2a---(*)$
it is also clear that $b=1$ is a solution of (*)
this gives $a=\pm 1$

 

[anemone]

Spoiler

it is clear that $a=0$ is a solution
let :$a^2=b, 2^a=c,$ we have :
$2bc-2b+2=2c-2^ba+2a---(*)$
it is also clear that $b=1$ is a solution of (*)
this gives $a=\pm 1$

Thanks for participating, Albert! :D

But, can you show us more explicitly how those three are the only solutions to the problem?

 

[Albert1]

Spoiler

it is clear that $a=0$ is a solution
let :$a^2=b, 2^a=c,$ we have :
$2bc-2b+2=2c-2^ba+2a---(*)$
it is also clear that $b=1$ is a solution of (*)
this gives $a=\pm 1$

continue:

Spoiler

it is clear that $a=0$ is a solution
let :$a^2=b, 2^a=c,$ we have :
$2bc-2b+2=2c-2^ba+2a---(*)$
it is also clear that $b=1$ is a solution of (*)
this gives $a=\pm 1$
now let $b=1+x>0$
and (*) becomes:
$x(2^a-1)=-a(2^x-1)$
or:
$x(2^\sqrt{1+x}-1)=-\sqrt{1+x}(2^x-1)----(**)$
(with 1+x>0)
if $x>0$ then the left side of (**)>0 ,but the right side of (**)<0---(1) and:
if $x<0$ then the left side of (**)<0,but the right side of (**)>0---(2)
from (1)(2):
$x=0,b=1, and, \,\, a=\pm 1$
$\therefore a=0,\pm 1$ those three are the only solutions to this problem

 

[anemone]

Thanks for your continued effort for submitting a complete solution, Albert! :)

I guess the second interval of $x$ as you stated in

if $x<0$ then the left side of (**)<0,but the right side of (**)>0

it should be $-1<x<0$ instead. :D

I will post the other solution that I want to share with MHB later.

 

[Albert1]

Thanks for your continued effort for submitting a complete solution, Albert! :)

I guess the second interval of $x$ as you stated in

it should be $-1<x<0$ instead. :D

I will post the other solution that I want to share with MHB later.

if x<0 ,for x+1>0, it is clear that x>-1