I Love Mr.Green

For discussing Olympiad Level Number Theory problems
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zadid xcalibured
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I Love Mr.Green

Unread post by zadid xcalibured » Thu Apr 11, 2013 9:47 pm

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

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Tahmid Hasan
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Re: I Love Mr.Green

Unread post by Tahmid Hasan » Thu Apr 11, 2013 11:59 pm

zadid xcalibured wrote:$a,b \in N$ such that $\forall n \in N_0$ ,$2^{n}a+b$ is a perfect square.Prove that $a=0$. :mrgreen:
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 :?
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*Mahi*
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Re: I Love Mr.Green

Unread post by *Mahi* » Fri Apr 12, 2013 1:30 am

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.)
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zadid xcalibured
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Re: I Love Mr.Green

Unread post by zadid xcalibured » Fri Apr 12, 2013 1:38 am

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

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Re: I Love Mr.Green

Unread post by *Mahi* » Fri Apr 12, 2013 2:25 am

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]
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zadid xcalibured
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Re: I Love Mr.Green

Unread post by zadid xcalibured » Fri Apr 12, 2013 11:29 pm

My solution is same. :mrgreen: :mrgreen: :mrgreen:

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