## I Love Mr.Green

- zadid xcalibured
**Posts:**217**Joined:**Thu Oct 27, 2011 11:04 am**Location:**mymensingh

### I Love Mr.Green

$a,b \in \mathbb N_0$ such that $\forall n \in \mathbb N_0$ ,$2^{n}a+b$ is a perfect square.Prove that $a=0$.

- Tahmid Hasan
**Posts:**665**Joined:**Thu Dec 09, 2010 5:34 pm**Location:**Khulna,Bangladesh.

### Re: I Love Mr.Green

I guess you meant $a,b \in \mathbb{N}_0$.zadid xcalibured wrote:$a,b \in N$ such that $\forall n \in N_0$ ,$2^{n}a+b$ is a perfect square.Prove that $a=0$.

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

বড় ভালবাসি তোমায়,মা

### Re: I Love Mr.Green

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.)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 $a=0$.

Please read Forum Guide and Rules before you post.

Use $L^AT_EX$, It makes our work a lot easier!

Nur Muhammad Shafiullah | Mahi

Use $L^AT_EX$, It makes our work a lot easier!

Nur Muhammad Shafiullah | Mahi

- zadid xcalibured
**Posts:**217**Joined:**Thu Oct 27, 2011 11:04 am**Location:**mymensingh

### Re: I Love Mr.Green

Well, I meant $a,b \in N_0$. Sorry for the typo.

I was searching for a title.But i couldn't find anything fitting with this problem.And i don't know the source either.So i gave this title on a whim.Because my forum favourite emo is mrgreen.

I was searching for a title.But i couldn't find anything fitting with this problem.And i don't know the source either.So i gave this title on a whim.Because my forum favourite emo is mrgreen.

### Re: I Love Mr.Green

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]

$2^na+b$ 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]

Please read Forum Guide and Rules before you post.

Use $L^AT_EX$, It makes our work a lot easier!

Nur Muhammad Shafiullah | Mahi

Use $L^AT_EX$, It makes our work a lot easier!

Nur Muhammad Shafiullah | Mahi

- zadid xcalibured
**Posts:**217**Joined:**Thu Oct 27, 2011 11:04 am**Location:**mymensingh

### Re: I Love Mr.Green

My solution is same.