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Fatin Farhan wrote: $$y^2-x^2=8$$
So, $$(y+x)=4,8$$ and $$(y-x)=2,1$$

There's a bug in the solution. $y$ and $x$ need not be integers. Only $N$ needs to be a positive integer. $\sqrt{1+8N}$ and $\sqrt{9+8N}$ may be rational, or, irrational.

Fatin Farhan wrote: $$\binom{n}{n-1}n=3n$$
So, $$n=3$$

Could you explain a bit more?? How did you get this equation??
And instead of using $*$ you can us the command "\times", if you want to write "$\times$".

Fatin Farhan wrote:8.
$$\binom{n}{n-1}n=3n$$
So, $$n=3$$

This could be written when n is variable , not necessarily true for constant $n$ . (Though answer is correct)
My solution to problem 8

$2n^k+3n=(n+1)^n-1=n [ (n+1)^{n-1}+(n+1)^{n-2}+..............(n+1)^1+1 ]$
$2n^k+3n \equiv n[(n+1)^{n-1}+(n+1)^{n-2}+..............(n+1)^1+1] \equiv n[(n-1)+1] \equiv 0$ ($mod$ $n^2$)

Hence $n^2|2n^k+3n \Rightarrow n|2n^{k-1}+3 $ , $\therefore n|3 , n\not=1 , n=3$
$n=3$ gives $k=3$.

I got stuck in 6 any hint ?

here are some non-geo probs ...

9. Find prime numbers $p$ such that $p+2$ and $p^2+2p-8$ are prime also .

10. $d$ and $d$ $'$ are natural number (where $d$ $'$ $>d$ ) and both are divisors of $n$ . Prove that $d$ $'$ $\displaystyle > d+\frac{d^2}{n}$

Source: [9]Albania NMO 2012, [10]Russia 2011

Solution of problem $5$:
Let there exists an positive integer $a$ such that
$\frac{1+\sqrt{1+8n}}{2}<a<\frac{1+\sqrt{9+8n}}{2}$
or $\sqrt{1+8n}<2a-1<\sqrt{9+8n}$
or $1+8n<(2a-1)^{2}<9+8n$
or $1+8n<4a^{2}-4a+1<9+8n$
or $8n<4a^{2}-4a<8+8n$
or $n<\frac{a(a-1)}{2}<1+n$
obviously $\frac{a(a-1)}{2}$ is a integer.But there is no integer between $n$ and $n+1$.So no such $a$ exists.

Solution to 9.
$$p^2+2p-8=(p+1)^2-9$$
$$=(p+1-3)(p+1+3)=(p+4)(p-2)$$
but it is a prime.
So,
$$p-2=1$$
$$\boxed{p=3}$$.
Now if $$p=3$$ both $$p+2,p^2+2p-8$$ are prime.

i have solved problem 9 in another ways;
$p,p+1,p+2$ -among these three integers one is divisible by 3.
if $3\mid p+2$ then p becomes 1. but given p is a prime. so it is impossible.
if $p\mid p+1\Leftrightarrow p+1\equiv 0 (mod 3)$
$\therefore p\equiv 2 (mod 3)\Leftrightarrow p^{2}\equiv 1(mod3)$
$\therefore 2p\equiv 1(mod3)$
so,$p^{2}+2p\equiv 2 (mod3)\Leftrightarrow p^{2}+2p-8\equiv -6(mod3)\Leftrightarrow p^{2}+2p-8\equiv 0(mod3)$ .but given $p^{2}+2p-8$ is also a prime. so it is also impossible.

last case 3 must divides p. but p is a prime. so $p=3$ [ans] imply that p is a prime, 3 divides p and $p+2;p^{2}+2p-8$ are both prime

problem 10 was stated wrong , now it is edited .

Hints for Geo Problems:
Problem $4$:Prove $A$ is the circumcenter of $\bigtriangleup BCD$
Problem $6$:Let $P=\bigodot ABC\cap \bigodot BED$.Then prove that $P,H,L$ and $P,G,M$ are collinear.

sowmitra wrote: $[6]$(APMO'08)
$\Gamma$ is the circum-circle of $\triangle ABC$. A circle passing through $A$ and $C$ meets $BC$ and $BA$ at $D$ and $E$ respectively. $AD$ and $CE$ intersect $\Gamma$ again at $G$ and $H$. The tangents of $\Gamma$ at $A$ and $C$ meet $DE$ at $L$ and $M$. Prove that, the intersection of $LH$ and $MG$ lies on $$\Gamma$$.

First assume $DE$ and $AC$ intersect. Let $DE \cap AC=K$. Let $AK \cap \odot ABC=N$.
Let $CC \cap GN=M'$. Applying Pascal on $CCAGNB$ we get $M',K,D$ are collinear, Hence $M'$ lies on $DK$ meaning $DE$.
So $M=M' \Rightarrow N,G,M$ are collinear. Similarly we can prove $N,H,L$ are collinear. So $LH$ and $MG$ intersect on $\odot ABC$.
If $DE \parallel AC$, do the same procedure with $DE \cap AC=\infty$.

Tahmid Hasan wrote: Let $A$$K \cap \odot ABC=N$.

I think it will be $B$$K \cap \odot ABC=N$.