### Re: Secondary and Higher Secondary Marathon

Posted:

**Thu Jan 31, 2013 12:45 pm**$n^2-n=n(n-1) \equiv 0 (mod10^5)$

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Posted: **Thu Jan 31, 2013 12:45 pm**

$n^2-n=n(n-1) \equiv 0 (mod10^5)$

Posted: **Thu Jan 31, 2013 1:58 pm**

Well now the full solution : 35

So we have two cases

Case 1 :

$n=5^5a, n-1=2^5b $

subtract . You will get a diophentine equation $5^5a-2^5b=1$

It has a solution like $(a,b)=(-3+32t,-293+3125t) \forall t \in \mathbb {Z}$

Its easy to verify that only $t=1$ is valid here .

So we get $n=90625$ .

Case 2 :

$n=2^5a, n-1=5^5b$

which lead us to the diophentine equation $2^5a-5^5b=1$

It has a solution like $(a,b)=(293+3125t,3+32t) \forall t \in \mathbb {Z}$

Its easy to verify that there is no valid $t$ in favour .

Note that $(n,n-1)=1$Nadim Ul Abrar wrote:$n^2-n=n(n-1) \equiv 0 (mod10^5)$

So we have two cases

Case 1 :

$n=5^5a, n-1=2^5b $

subtract . You will get a diophentine equation $5^5a-2^5b=1$

It has a solution like $(a,b)=(-3+32t,-293+3125t) \forall t \in \mathbb {Z}$

Its easy to verify that only $t=1$ is valid here .

So we get $n=90625$ .

Case 2 :

$n=2^5a, n-1=5^5b$

which lead us to the diophentine equation $2^5a-5^5b=1$

It has a solution like $(a,b)=(293+3125t,3+32t) \forall t \in \mathbb {Z}$

Its easy to verify that there is no valid $t$ in favour .

Posted: **Thu Jan 31, 2013 3:05 pm**

Problem $\boxed {36}$ (A beautiful problem)

Suppose $ABCD$ is a rectangle and $P,Q,R,S$ are points on the sides $AB, BC, CD. DA$ respectively. Show that

$PQ+QR+RS+SP \geq 2AC$

source : AoPs

Suppose $ABCD$ is a rectangle and $P,Q,R,S$ are points on the sides $AB, BC, CD. DA$ respectively. Show that

$PQ+QR+RS+SP \geq 2AC$

source : AoPs

Posted: **Thu Jan 31, 2013 6:23 pm**

Let $P_1,P_2$ be reflection of $P$ over $AD,BC$ respectively. Let $P_3$ be the reflection of $P_1$ over $CD$.Nadim Ul Abrar wrote:Problem $\boxed {36}$ (A beautiful problem)

Suppose $ABCD$ is a rectangle and $P,Q,R,S$ are points on the sides $AB, BC, CD. DA$ respectively. Show that

$PQ+QR+RS+SP \geq 2AC$

source : AoPs

Now $PQ+QR+RS+SP=(P_2Q+QR)+(RS+SP_1) \ge P_2R+P_1R=P_2R+P_3R \ge P_2P_3$

$=\sqrt{P_1P_3^2+P_1P_2^2}=\sqrt{(2AD)^2+(2AB)^2}=2AC$.

Someone else post the next problem.

Posted: **Thu Jan 31, 2013 7:09 pm**

WoW . Waht a Co-incidence . I did post quite same solution there .

Posted: **Thu Jan 31, 2013 8:11 pm**

Great minds think alike!Nadim Ul Abrar wrote:WoW . Waht a Co-incidence . I did post quite same solution there .

Well, actually once Pritom showed me a transformation problem involving hexagons and the solution he made used same kinds of reflections I used here. So you may call it a copied idea

Posted: **Fri Feb 01, 2013 9:54 pm**

Probelm $37$: $\mathbb{N}$ is the set of positive integers and $a\in\mathbb{N}$. We know that for every $n\in\mathbb{N}$, $4(a^n+1)$ is a perfect cube. Prove that $a=1$.

Source: Iran NMO-2012-4.

Note: In BdMO Summer Camp-2012, a similar problem was given in the Number Theory problem set.

Zubaer vai gave a 'cruxy' solution, but unfortunately I don't remember it. So I solved it with brute force.

Does anybody remember?

Source: Iran NMO-2012-4.

Note: In BdMO Summer Camp-2012, a similar problem was given in the Number Theory problem set.

Zubaer vai gave a 'cruxy' solution, but unfortunately I don't remember it. So I solved it with brute force.

Does anybody remember?

Posted: **Sat Feb 02, 2013 1:57 pm**

L.T.E. finishes it just in two lines. But don't appreciate this kind of proofs now. So I looked for other proofs but went in vain. I am very unhappy with my L.T.E. based proof. If anybody can find any other proof, please post it here. @Tahmid vai, please post your proof and if you can still remember Zubayer vai's nice proof, please post it, too.Tahmid Hasan wrote:Probelm $37$: $\mathbb{N}$ is the set of positive integers and $a\in\mathbb{N}$. We know that for every $n\in\mathbb{N}$, $4(a^n+1)$ is a perfect cube. Prove that $a=1$.

Source: Iran NMO-2012-4.

Note: In BdMO Summer Camp-2012, a similar problem was given in the Number Theory problem set.

Zubaer vai gave a 'cruxy' solution, but unfortunately I don't remember it. So I solved it with brute force.

Does anybody remember?

Divide both sides of the equation by four and define, $a^2+1=2z^3$

Taking $\pmod {4}$ shows $z$ is odd. If $z=1$, we are done, if not, then let $p$ be a prime divisor of $z$. Note that $p$ is odd and co-prime to $a^2$. Suppose $p^{3c}||a^2+1$. So $p^{3c+1}||4\big ((a^2)^p+1^p\big )=4(a^{2p}+1)\Longrightarrow \quad$ it is not a perfect cube. A contradiction, so done.

Posted: **Sun Feb 03, 2013 8:55 pm**

There's a typo, it's from Iran NMO-2008Tahmid Hasan wrote:Probelm $37$: $\mathbb{N}$ is the set of positive integers and $a\in\mathbb{N}$. We know that for every $n\in\mathbb{N}$, $4(a^n+1)$ is a perfect cube. Prove that $a=1$.

Source: Iran NMO-2012-4.

Note: In BdMO Summer Camp-2012, a similar problem was given in the Number Theory problem set.

Zubaer vai gave a 'cruxy' solution, but unfortunately I don't remember it. So I solved it with brute force.

Does anybody remember?

Sorry, I don't but I remember Mahi and Nadim vai solved it using ring.Phelembac Adib Hasan wrote:@Tahmid vai, please post your proof and if you can still remember Zubayer vai's nice proof, please post it, too.

My solution: Note that if $a=1,4(a^n+1)=8=2^3$.

Now assume $a>1$.

$4(a^9+1),4(a^3+1)$ are both perfect cubes, so their quotient $a^6-a^3+1$ is a perfect cube too.

$\forall a>1,a^3-1>0 \Rightarrow a^6-a^3+1<(a^2)^3$

So $a^6-a^3+1 \le (a^2-1)^3 \Rightarrow a^2(3a^2-a-3)+2 \le 0$.

But $\forall a>1, 3a^2>a+3$, so we have a contradiction.

Posted: **Tue Feb 05, 2013 9:14 pm**

Problem 38:

$3^x+7^y=n^2$

how many integer solutions for $(x,y)$ are there?

$3^x+7^y=n^2$

how many integer solutions for $(x,y)$ are there?