## Dhaka Secondary 2010/6

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**don't post problems (by starting a topic)**in the "Secondary: Solved" forum. This forum is only for showcasing the problems for the convenience of the users. You can post the problems in the main Divisional Math Olympiad forum. Later we shall move that topic with proper formatting, and post in the resource section.

### Dhaka Secondary 2010/6

For how many prime numbers $N$ for which $N+1$ is a perfect square.

- leonardo shawon
**Posts:**169**Joined:**Sat Jan 01, 2011 4:59 pm**Location:**Dhaka

### Re: Dhaka Secondary 2010/6

There is an only number which is 3 and 3+1 is a sqare .

- leonardo shawon
**Posts:**169**Joined:**Sat Jan 01, 2011 4:59 pm**Location:**Dhaka

### Re: Dhaka Secondary 2010/6

actually the question was HOW Many Numbers! So i wrote the answer.. Thats there r 1 number.

and yes. Only number is 3. And 3+1=4 is a perfect square number.

and yes. Only number is 3. And 3+1=4 is a perfect square number.

Ibtehaz Shawon

BRAC University.

BRAC University.

*long way to go .....*-
**Posts:**7**Joined:**Wed Jan 26, 2011 9:36 am**Location:**Sylhet-
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### Re: Dhaka Secondary 2010/6

Yaap . Ami bistarito vabe bollam .

### Re: Dhaka Secondary 2010/6

We should prove that 3 is only such prime.

If $N+1=k^2$ then $N=k^2-1 \Rightarrow N=(k-1)(k+1)$ as $N$ is prime, $k-1=1\ \ \therefore k=2$

If $N+1=k^2$ then $N=k^2-1 \Rightarrow N=(k-1)(k+1)$ as $N$ is prime, $k-1=1\ \ \therefore k=2$

Every logical solution to a problem has its own beauty.

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**Important**: Please make sure that you have read about the**Rules, Posting Permissions and Forum Language**)- leonardo shawon
**Posts:**169**Joined:**Sat Jan 01, 2011 4:59 pm**Location:**Dhaka

### Re: Dhaka Secondary 2010/6

how did it become?Zzzz wrote:We should prove that 3 is only such prime.

If $N+1=k^2$ then $N=k^2-1 \Rightarrow N=(k-1)(k+1)$ as $N$ is prime, $k-1=1\ \ \therefore k=2$

$(k-1)(k+1)=N as N is a prime, so (k-1)=1 . . . ?

Ibtehaz Shawon

BRAC University.

BRAC University.

*long way to go .....*### Re: Dhaka Secondary 2010/6

k+1 and k-1 both are factors of a prime number. Each prime number has only two factors - 1 and the prime itself. So k-1=1.leonardo shawon wrote:how did it become?Zzzz wrote:We should prove that 3 is only such prime.

If $N+1=k^2$ then $N=k^2-1 \Rightarrow N=(k-1)(k+1)$ as $N$ is prime, $k-1=1\ \ \therefore k=2$

$(k-1)(k+1)=N as N is a prime, so (k-1)=1 . . . ?

Every logical solution to a problem has its own beauty.

(

(

**Important**: Please make sure that you have read about the**Rules, Posting Permissions and Forum Language**)