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Algebra

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Joined:	23 Dec 2006
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4 Feb 2007 12:52:55 IST

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526

Celebrate my CENTURY

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This is the 100th message that I am posting. So in order to celebrate this ocassion, I am giving a problem which I believe, everyone will like.

Suppose a,b,c are integers with a > 1, & p is a prime number. Show that if ax^2 + bx + c is equal to p for two distinct integral values of x, then ax^2 + bx + c cannot be equal to 2p for any integral value of x.

I enjoyed this sum & I hope others will enjoy this too.

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Comments (4)

amaron

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Joined: 26 Jan 2007

Posts: 156

4 Feb 2007 13:35:45 IST

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let ax² + bx +c=p have roots

(are integers)

=c-p/a
c-p=a

(1)

let ax² + bx +c=2p have roots(j,k are integers)
c-2p=ajk (2)

in (1) subtract p on both sides and substitute in (2)
a

-p=ajk

implies
a(

-jk)=p
as a>1 (

-jk)is an integer and p is prime

this is not possible.

that solves the problem!

AAKRITI

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Joined: 31 Dec 2006

Posts: 291

4 Feb 2007 13:39:01 IST

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yes i also wanted to give similar solution as the above one

from the 1st equation (c-p)/a belongs to I as the product of 2 integers is integer

then from 2nd (c-2p)/a belongs to I

i.e. (c-p)/a -p/a is integer, this cant be true as p is not divisible by a

AAKRITI

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Joined: 31 Dec 2006

Posts: 291

4 Feb 2007 15:34:41 IST

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i would like to point out one thing here, that a should not be equal to p otherwise product of roots for the 2nd equation will become (c-2p)/p = (c-p)/p -1, now since (c-p)/p is aleady an integer, (c-2p)/p also becomes integer.

(this is my 100th post)

AAKRITI

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Joined: 31 Dec 2006

Posts: 291

5 Feb 2007 22:12:36 IST

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atleast tell me what way you proved it

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