Math Is Fun Forum

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Is there a polynomial?

Hi;

I am unable to prove this one way or another:

Can there be a polynomial f(x), that has integer coefficients and f(2) = 1 and f(6) = 4?

This popped up on another forum and I am stumped.

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In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: Is there a polynomial?

Here's an easier one.
Can there be a polynomial g(x) with integer coefficients and g(0) = 0 and g(4) = 3?

Solve that and you're halfway to solving yours.

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Why did the vector cross the road?
It wanted to be normal.

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Is there a polynomial?

Hi mathsyperson;

Can there be a polynomial g(x) with integer coefficients and g(0) = 0 and g(4) = 3?

Using this method here at the last page of the PDF problem # 16;

http://docs.google.com/viewer?a=v&q=cac … pU0X0EDHBA

Starting with the general polynomial:

Then P(r) - P(s) is:

We can pull out a factor (r-s) leaving Q(r,s) which is an integer.

The problem states P(0) = 0 and P(4) = 3.

Just plug into the last formula:

Since 3 cannot be equal to 4 times an integer we have a contradiction therefore no poly like that exists.

Same method proves mine cannot exist either.

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In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Is there a polynomial?

Same method proves mine cannot exist either.

Perhaps, but there is no reason to do all that work again.  You've in fact just proven that your polynomial doesn't exist in post #3.  Because if the polynomial in post #1 did exist, then the polynomial in post #3 would have to exist (vertical and horizontal shift).  That's a contradiction right there.

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

soroban

Member

Registered: 2007-03-09

Posts: 452

Re: Is there a polynomial?

. .

.

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: Is there a polynomial?

Oh, I like that a lot. f(2) and f(6) must both be [even number] + k, and so they must have the same parity. We're asking for a function where their parities are different, so such a function doesn't exist.

Brilliantly simple.

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Why did the vector cross the road?
It wanted to be normal.

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Is there a polynomial?

Hi mathsyperson, Ricky and soroban;

Thanks for providing more than I knew of about this problem. Happy holiday!

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In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.