If $a,b\in \mathbb {N}_0$, show that $\sqrt {a}+\sqrt {b}$ is rational iff both of $a$ and $b$ are

Source: Self-made.

- Phlembac Adib Hasan
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Problem 4: (Posted before)

If $a,b\in \mathbb {N}_0$, show that $\sqrt {a}+\sqrt {b}$ is rational iff both of $a$ and $b$ are~~integers~~ perfect squares.

Source: Self-made.

If $a,b\in \mathbb {N}_0$, show that $\sqrt {a}+\sqrt {b}$ is rational iff both of $a$ and $b$ are

Source: Self-made.

Last edited by Phlembac Adib Hasan on Tue Nov 13, 2012 7:22 pm, edited 2 times in total.

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- Fahim Shahriar
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Phlembac Adib Hasan wrote: Problem 4: (Posted before)

If $a,b\in \mathbb {N}_0$, show that $\sqrt {a}+\sqrt {b}$ is rational iff both of $a$ and $b$ are integers.

Source: Self-made.

Name: **Fahim Shahriar Shakkhor**

Notre Dame College

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- Phlembac Adib Hasan
**Posts:**1016**Joined:**Tue Nov 22, 2011 7:49 pm**Location:**127.0.0.1-
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Sorry, for the typo. Edited now.Fahim Shahriar wrote:Phlembac Adib Hasan wrote: Problem 4: (Posted before)

If $a,b\in \mathbb {N}_0$, show that $\sqrt {a}+\sqrt {b}$ is rational iff both of $a$ and $b$ are integers.

Source: Self-made.

Is it true ?

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First if $a,b$ are both perfect squares,then $\sqrt a+\sqrt b$ must be an integer,therefore,rational. So the "only if" case is true.

Now let us assume that $\sqrt a+\sqrt b=\frac{m}{n}$. Now squaring both sides, we get

$a+2\sqrt a\cdot \sqrt b+b=\frac{m^2}{n^2}$.

This implies $2\sqrt a\cdot \sqrt b=\frac{m^2-n^2(a+b)}{n^2}\Rightarrow 2n^2\sqrt a\cdot \sqrt b=m^2-n^2(a+b)$.

So we must have $\sqrt {ab}\in \mathbb N$. Thus we must have $\sqrt {ab}=\frac{x}{y}$. Thus $ab=\frac{x^2}{y^2}$.

Now, $\sqrt a+\sqrt b=\sqrt a+\frac{x}{y\sqrt a}=\frac{x+ay}{y\sqrt a}=\frac{m}{n}$. Thus $\sqrt a$ must be rational.

So for some $u,v\in \mathbb Z$ we have that $a=\frac{u^2}{v^2}$. But we have that $a\in \mathbb N_{0}$. So $a$ must be a square integer i.e. $v|u$ . Similar steps for $b$.

Now let us assume that $\sqrt a+\sqrt b=\frac{m}{n}$. Now squaring both sides, we get

$a+2\sqrt a\cdot \sqrt b+b=\frac{m^2}{n^2}$.

This implies $2\sqrt a\cdot \sqrt b=\frac{m^2-n^2(a+b)}{n^2}\Rightarrow 2n^2\sqrt a\cdot \sqrt b=m^2-n^2(a+b)$.

So we must have $\sqrt {ab}\in \mathbb N$. Thus we must have $\sqrt {ab}=\frac{x}{y}$. Thus $ab=\frac{x^2}{y^2}$.

Now, $\sqrt a+\sqrt b=\sqrt a+\frac{x}{y\sqrt a}=\frac{x+ay}{y\sqrt a}=\frac{m}{n}$. Thus $\sqrt a$ must be rational.

So for some $u,v\in \mathbb Z$ we have that $a=\frac{u^2}{v^2}$. But we have that $a\in \mathbb N_{0}$. So $a$ must be a square integer i.e. $v|u$ . Similar steps for $b$.

$\color{blue}{\textit{To}} \color{red}{\textit{ problems }} \color{blue}{\textit{I am encountering with-}} \color{green}{\textit{AVADA KEDAVRA!}}$

- Fahim Shahriar
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The sum of three numbers is $6$, the sum of their squares is $8$, and the sum of their cubes is $5$. What is the sum of their fourth powers?

Source: অলিম্পিয়াড সমগ্র বই

Name: **Fahim Shahriar Shakkhor**

Notre Dame College

Notre Dame College

- Phlembac Adib Hasan
**Posts:**1016**Joined:**Tue Nov 22, 2011 7:49 pm**Location:**127.0.0.1-
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assume the numbers are $a,b,c$Fahim Shahriar wrote:Problem 5:

The sum of three numbers is $6$, the sum of their squares is $8$, and the sum of their cubes is $5$. What is the sum of their fourth powers?

Source: অলিম্পিয়াড সমগ্র বই

So $ab+bc+ca=1/2[(a+b+c)^2-a^2-b^2-c^2]=14$

So $a^4+b^4+c^4=0$

Let me know whether I am right.

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- Fahim Shahriar
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Phlembac Adib Hasan wrote:Let me know whether I am right.

Yes. You're right.

Who is going to post the next one?

Name: **Fahim Shahriar Shakkhor**

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- Phlembac Adib Hasan
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Source: http://www.brilliantscholars.com/assess ... ry/114744/

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

Sophie Jermain kills it.Phlembac Adib Hasan wrote:Problem 6:Find all integers $n$ such that $n^4+4$ is a prime.

Source: http://www.brilliantscholars.com/assess ... ry/114744/

I'm posting a new problem.

Problem $7$:The interior bisector of $\angle A$ from $\triangle ABC$ intersects the side $BC$ and the circumcircle of $\triangle ABC$ at $D;M$, respectively. Let $\omega$ be a circle with center $M$ and radius $MB$. A line passing through $D$, intersects $\omega$ at $X;Y$. Prove that $AD$ bisects $\angle XAY$.

Source: Iran NMO-2004-5

বড় ভালবাসি তোমায়,মা

i couldn't get that ..... i did problem 6 with factorising which results $1,-1$. i think this is general way.Tahmid Hasan wrote: Sophie Jermain kills it.

$\angle BAM=\angle CAM \Rightarrow \angle BCM=\angle CBM$ , $BM=CM$.so $C$ point lies on the $\omega$ circle .

now , $XD.DY=BD.DC=AD.DM$

$\therefore \displaystyle \frac{XD}{AD}=\frac{DM}{DY}$ , $\Delta AXD \sim \Delta DMY$. $\angle XAD=\angle MYD$ . $\therefore A,X,M,Y $ are cyclic.

$MX=MY \Rightarrow \angle XYM=\angle MXY$

$\angle XAM=\angle XYM=\angle MXY=\angle MAY$ ,

$AD$ bisects $\angle XAY$.

someone post the next problem.