If we denote $f(n)$ the desired number, then CRT implies $f(mn)=f(m)f(n)$ whenever $(m;n)=1$. So it suffices to find $f(n)$ when $n$ is a prime power. It's quite trivial that there are $2$ solutions for $x $ if $n $ is a prime power. So, the answer is $2^z $ where $z $ is the number of distinct prim...

- Thu Jun 22, 2017 6:51 am
- Forum: Number Theory
- Topic: $x^2 \equiv x (mod n)$
- Replies:
**1** - Views:
**89**

We rearrange the equation as, $n^5+n^4+1= 7^m $. Observe that $n^5+n^4+1=(n^2+n+1)(n^3-n+1)$. So, each of $n^2+n+1$ and $n^3-n+1$ is $7^x $ for some integer $x \geq 0$. But using the euclidean algorithm, we find that $gcd(n^2+n+1, n^3-n+1)$ divides $7$. So we consider the $4$ equations $n^2+n+1=1$ $...

- Thu Jun 22, 2017 6:20 am
- Forum: Number Theory
- Topic: n^5+n^4
- Replies:
**1** - Views:
**52**

The left hand side is even, so $ x^4+y^2$ is even, and $ x, y$ has the same parity. However, if $ x, y$ are odd, we have that $ 4$ divides $ 7^x-3^y$, but $ 4$ doesn't divide $ x^4+y^2$. So $ x, y$ are even, and let's say $ x=2x_1,y=2y_1$. You'll get $ (7^{x_1}-3^{y_1})(7^{x_1}+3^{y_1})$ divides $ 4...

- Thu Jun 22, 2017 5:42 am
- Forum: Number Theory
- Topic: Divisibility... with x,y
- Replies:
**1** - Views:
**57**

Ugly problem tbh. Let us assign $\angle BAC = 4a$. Then quick angle chases give us, $\angle CAK = a $, $\angle DCI = 45-a $, $\angle CEK = 3a $. We can apply sine law on $\triangle IEK $ and $\triangle CEK $ to get $ \frac{IK}{KC} = \frac{sin(45)*sin (45-a)}{sin(3a)*sin(90-2a)}$ ........(relation 1)...

- Thu Jun 22, 2017 4:53 am
- Forum: Geometry
- Topic: IMO 2009 SL(G-1)
- Replies:
**1** - Views:
**64**

Problem $23$

Let $a,b,c$ be positive integers. Prove that it is impossible to have all of the three numbers $a^2+b+c,b^2+c+a,c^2+a+b$ to be perfect squares.

Let $a,b,c$ be positive integers. Prove that it is impossible to have all of the three numbers $a^2+b+c,b^2+c+a,c^2+a+b$ to be perfect squares.

- Wed Jun 21, 2017 5:26 pm
- Forum: Junior Level
- Topic: Beginner's Marathon
- Replies:
**63** - Views:
**2132**

They start, from Mirpur, one with the bike($ A$) and the other($ B$) on foot, and from Lalmatia($ C$) on foot of course. After one hour $ A$ will give the bike to $ C$, and $ B$ will stop, and wait for the bike. $ A$ will reach Lamatia after 2 hours. $ C$ rides the bike until he meets $ B$ , and giv...

- Tue Jun 20, 2017 10:39 pm
- Forum: Junior Level
- Topic: Beginner's Marathon
- Replies:
**63** - Views:
**2132**

I edited the post.

- Thu Jun 15, 2017 2:47 pm
- Forum: Junior Level
- Topic: Beginner's Marathon
- Replies:
**63** - Views:
**2132**

Problem $22$ The distance between Mirpur and Lalmatia is $24$ km. Two of three friends need to reach Lalmatia from Mirpur and another friend wants to reach Mirpur from Lalmatia. They have only one bike, which is initially in Mirpur. Each guy may go on foot (with velocity at most $6$ kmph) or on a bi...

- Thu Jun 15, 2017 2:24 pm
- Forum: Junior Level
- Topic: Beginner's Marathon
- Replies:
**63** - Views:
**2132**

OKAY PEOPLE!! Time to make this thread hot again. Problem $18$ You and your spouse invited $10$ couples to a party. In that party, some of them shook hands with others. But no one shook hands with their spouse. After the party, you asked everyone(including your wife) how many person they shook hands...

- Wed Jun 07, 2017 3:19 pm
- Forum: Junior Level
- Topic: Beginner's Marathon
- Replies:
**63** - Views:
**2132**

No, there are solutions. You have to consider the case where $k+4-4m^2=0$ or $n^2+k-m^2=0$.

- Wed Jun 07, 2017 2:37 pm
- Forum: Divisional Math Olympiad
- Topic: 2017 Regional no.9 Dhaka
- Replies:
**8** - Views:
**197**