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

Algebra

Forum Expert

Joined:	28 Feb 2007
Post:	2173

25 Jan 2008 09:03:07 IST

0 People liked this

482

One more polynomial problem

None

Konichiwa stay away from this one! (thats a clue

)

p(x) is a polynomial with integer coefficients. For some positive integer c, none of p(1), p(2), ... , p(c) are divisible by c. Prove that p(b) is not zero for any integer b.

Share this article on:

Comments (3)

abhishek sinha

Forum Expert

Joined: 18 Dec 2007

Posts: 926

25 Jan 2008 21:56:30 IST

Like 4 people liked this

If possible let p( b ) = 0 for an integer b .

case - I ( b< 1 )

Now , by division algorithm , we may write

p ( x ) = ( x - b ) Q ( x ) ( Q ( x ) being a polynomial with integer coeff s ) ..............( 1 )

Now we have p ( 1 ) = ( 1 - b ) Q ( 1 )

p ( 2 ) = ( 2 - b ) Q ( 2 )

...............................................

p( c ) = ( c- b ) Q ( c )

Now see that all Q ( i ) s are integer ............ ( 2 )

& among ( 1 - b ) , ( 2- b ) , ............ ( c- b ) { They are c consecutive integers & all 0 or -ve ) at least one must be divisible by c ............ ( 3 )

So from ( 2 ) & ( 3 ) it follows that at least one of p( 1 ) , p( 2 ) ,............p( c )

must be divisible by c .

But according to the given condition this is simply wrong .

So our hypothesis is wrong ( Reductio ad absurdum !!!!!!! )

So p( b ) is not zero for any negative integer b . ( Proved.........1 )

case - II

If b >c we take the factor to be ( b - x ) & all of ( b- 1 ) ( b- 2 ) ,, ..... ( b-c ) are positve and consecutive integers so atleast one must be divisible by c .

So our hypothesis is wrong .

so p( b ) is not 0 for any b>c. ( proved ........ 2 )

case - III

Now we have to prove for 1 < b <c

we see that for this case there is a b < c for which p( b ) = 0 , so p( b ) is divisible by c which is again against the given condition .

again our hypothesis is wrong .

so p( b ) is non zero for 1 < b < c ( proved ......... 3 )

So we havecompleted the proof for all possible cases.

Sairam

Blazing goIITian

Joined: 14 Sep 2007

Posts: 573

25 Jan 2008 23:04:23 IST

Like 0 people liked this

post edited

Hari Shankar

Forum Expert

Joined: 28 Feb 2007

Posts: 2173

26 Jan 2008 09:26:03 IST

Like 2 people liked this

We again use the result that P(a) - P(b) is div by (a-b)

Now, if an integer N is div by c then N = kc

We must have (kc-c)| P(kc) - P(c) So if P(N) = 0, then P(c) is div by c which is a contradiction.

Now, suppose N is not a multiple of c. Then N = cd+r where o<r<c

Then N-r|P(N) - P(r) or cd|P(N)-P(r). So, if P(N) = 0, P(r) is divisible by c which is a contradiction as P(k) is not divisible by c for k<=c.
Hence P(x) = 0 has no solution in integers

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