Home » Ask & Discuss » Mathematics. » Algebra « Back to Discussion

Algebra

Forum Expert

Joined:	28 Feb 2007
Post:	2173

4 Jul 2008 19:11:46 IST

0 People liked this

760

For my good friends rahul, rudra panda & Co.

None

Since you guys are interested,

1. How many pairs of positive integers (a,b) exist such that gcd(a,b) = 1 and $\frac{a}{b} + \frac{14b}{9a}$ is an integer

2. Find all n $\in \mathbb{N}$ such that $\frac{(n+1)^2}{n+7}$ is also a natural number.

Share this article on:

Comments (9)

Rahul Duggal

Scorching goIITian

Joined: 10 May 2008

Posts: 289

4 Jul 2008 20:48:08 IST

Like 1 people liked this

Re:For my good friends rahul, rudra panda & Co.

Hari Shankar

Forum Expert

Joined: 28 Feb 2007

Posts: 2173

5 Jul 2008 09:24:27 IST

Like 2 people liked this

Good, you got almost all the solutions except n=29. Incidentally, I found this problem in some coaching material which is for for the Ramaiah Coaching Entrance! Coaching for an entrance for coaching for another entrance!! Life has got enormously complicated these days. The solution given was long-wiunded so I wanted to see how you guys fare.

The shortest solution is at its crux a simple application of remainder theorem.

(n+1)² = (n+7) I + r , where I is an integer. The remainder r is a constant and is easily evaluated using Bezout's Theorem by putting n = -7. We obtain r = 36.

Hence, $\frac{(n+1)^2}{n+7} = I + \frac{36}{n+7}$

The problem reduces to finding n such that n+7|36. This happens for n=2,5,11,29

Rahul Duggal

Scorching goIITian

Joined: 10 May 2008

Posts: 289

5 Jul 2008 09:55:38 IST

Like 0 people liked this

sir where have you used bezout's theorem and what does it state

Hari Shankar

Forum Expert

Joined: 28 Feb 2007

Posts: 2173

5 Jul 2008 10:15:37 IST

Like 1 people liked this

Bezout's Theorem is the application of the Remainder Theorem for the special case when the polynomial P(x) is divided by a linear polynomial (ax+b). The theorem states that the remainder is $P(-\frac{b}{a})$

I think it is familiar to you in the form that the remainder when a polynomial P(x) is divided by (x-a) is P(a).

Rahul Duggal

Scorching goIITian

Joined: 10 May 2008

Posts: 289

5 Jul 2008 15:54:59 IST

Like 1 people liked this

sir i could not find more solutions exept (1,3)

Rahul Duggal

Scorching goIITian

Joined: 10 May 2008

Posts: 289

6 Jul 2008 07:53:16 IST

Like 0 people liked this

sir can you tell me how to proceed furtherfor the first question

Hari Shankar

Forum Expert

Joined: 28 Feb 2007

Posts: 2173

6 Jul 2008 14:24:47 IST

Like 2 people liked this

Sorry, my post is long overdue.

Ok, you must first rewrite the equation as

9a^2+14b^2 = 9abn

Now, 14b² and 9abn are both divisible by b. Hence 9a² must also be divisible by b

Now since gcd(a,b) = 1, b divides 9a² means, b divides 9. Also, 9 must divide b². The only possibility is b = 9.

Similarly you conclude that a divides 14.

So a = 1,2, 7 or 14 and b = 3.

Plugging in these you see that all the pairs (1,3), (2,3), (7,3) and (14,3) are solutions and no other solutions exist.

Rahul Duggal

Scorching goIITian

Joined: 10 May 2008

Posts: 289

6 Jul 2008 14:33:05 IST

Like 0 people liked this

sir why cant we take b=9 as 9 should be divisible by b so 9/9=1

Hari Shankar

Forum Expert

Joined: 28 Feb 2007

Posts: 2173

6 Jul 2008 17:38:25 IST

Like 3 people liked this

That's a valid question which I have failed to address. Thank you for pointing it out.

Again go back to the equation

9a^2+14b^2 = 9abn

Suppose b = 9. Then you can divide by 9 throughout and you would get

a²+14x9 = 9an

a is not a multiple of 9, but the other terms are. So you obtain a contradiction.

Quick Reply

Reply

	Some HTML allowed.
	Keep your comments above the belt or risk having them deleted.
	Signup for a avatar to have your pictures show up by your comment
	If Members see a thread that violates the Posting Rules, bring it to the attention of the Moderator Team