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I Love Mr.Green
Posted: Thu Apr 11, 2013 9:47 pm
by zadid xcalibured
$a,b \in \mathbb N_0$ such that $\forall n \in \mathbb N_0$ ,$2^{n}a+b$ is a perfect square.Prove that $a=0$.
Re: I Love Mr.Green
Posted: Thu Apr 11, 2013 11:59 pm
by Tahmid Hasan
zadid xcalibured wrote:$a,b \in N$ such that $\forall n \in N_0$ ,$2^{n}a+b$ is a perfect square.Prove that $a=0$.
I guess you meant $a,b \in \mathbb{N}_0$.
If b=0, then $a,2a$ are both perfect square but $\frac {2a}{a}=2$, which is not a perfect square, so a contradiction.
So we assume $a,b \neq 0$. We define $x_n=2^na+b$.
$2x_n-x_{n+1}=b$, taking limit $n \rightarrow \infty$, we get $x_n=b$
which implies $a=0$.
Off-topic: What's with the title? I don't get it
Re: I Love Mr.Green
Posted: Fri Apr 12, 2013 1:30 am
by *Mahi*
Tahmid Hasan wrote:We define $x_n=2^na+b$.
$2x_n-x_{n+1}=b$, taking limit $n \rightarrow \infty$, we get $x_n=b$
which implies $a=0$.
This is wrong. You can't use this argument without proving $\{ x_n\}^\infty$ converges. (also another easier hint, you did not use $2^na+b$ a square at all, you just proved that any series with $x_n = 2^na+b $ has $a = 0$, which is obviously false.)
Re: I Love Mr.Green
Posted: Fri Apr 12, 2013 1:38 am
by zadid xcalibured
Re: I Love Mr.Green
Posted: Fri Apr 12, 2013 2:25 am
by *Mahi*
Forgot my previous solution
new one:
$2^na+b$ is a perfect square $\Rightarrow 4.2^na+ 4b$ is a perfect square.
Again $2^{n+2}a+b = 4.2^na + b$ is a perfect square.
Now, $ 4.2^na+ 4b > 4.2^na+ b$
As both of them are perfect squares, $ 4.2^na+ 4b \geq (\sqrt{4.2^na+ b}+1)^2 = 4.2^na+ b + 2 \sqrt{ 4.2^na+ b}+1$.
So, $3b \geq 2 \sqrt{ 4.2^na+ b}+1$
But this is true for all $n \in \mathbb N_0$.
If $a>0$ then the RHS is unbounded, so contradiction.
So, $a=0$[proved]
Re: I Love Mr.Green
Posted: Fri Apr 12, 2013 11:29 pm
by zadid xcalibured