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Algebra

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8 Aug 2008 19:33:26 IST

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Prove even number

None

Let P(x) = a_0 + a_1 x + a_2 x^2 +...+ a_n x^n

We are given $P(-1) = 0; P(\sqrt 2)$ is an integer

Prove that we can always find a k such that P(k) + a_k is an even number.

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abhishek sinha

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9 Aug 2008 18:27:38 IST

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Confusing !!

One should mention the field which a i's belong to ( i.e.the field of complex no or real no, or rational or integer )

also to be meaningful , it should also be mentioned that 0<=k<=n .

Hari Shankar

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9 Aug 2008 18:48:11 IST

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oops my bad! they are all integers

abhishek sinha

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9 Aug 2008 19:24:25 IST

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Then k= 0 will do ( 2 a0)

( again confusing !!!)

Hari Shankar

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9 Aug 2008 19:32:41 IST

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ok baba, k>0.

abhishek sinha

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9 Aug 2008 20:19:03 IST

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Then P( sqrt 2) = integer implies that

sqrt 2 *integer ( containing the odd a i's ) = integer

How is that possible ?

Hari Shankar

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9 Aug 2008 20:21:32 IST

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Thought it would be difficult.

So, I will give the solution. Please see http://www.mathlinks.ro/viewtopic.php?t=219274

This is a basic concept that if a+b $\sqrt 2$ = a rational number then, well, b = 0

Here, b = a_1+2a_3+4a_5+...

which means is a₁ an even number.

P(-1) = 0 means a_1+a_3+a_5+... = a_2+a_4+a_6+....

So that P(1) is also even. Thus the result that P(1)+a₁is always even.

Pretty easy if you can get it!

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