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12 May 2011 19:21:56 IST

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A polynomial f(x) has integer coefficients such that f(0) and f(1) are both odd numbers Prove that

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A polynomial f(x) has integer coefficients such that f(0) and f(1) are both odd numbers Prove that f(x) = 0 has no integer solutions.

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jagdish singh

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13 May 2011 10:37:45 IST

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abhishek sinha

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14 May 2011 18:49:17 IST

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If possible, suppose that f(x) has an integral root , let it be "b"

then we may write

f(x)=(x-b)*Q(x) (where Q(x) is a polynomial of degree one less than f(x) )

ow since f(x) is a polynomial with integer coeffiecients, it follows (by just multiplying out term by term, Q(x) is also a polynomial with integral coeffiecients) ..... (1)

So, we have

f(0)*f(1)=(-b)*(1-b)*Q(0)*Q(1) = odd number (since f(0) and f(1) are both odd)

also Q(0) and Q(1) are integers (by (1)) and (-b) and (1-b) are two consecutive integers, hence there product is always even.

Hence (-b)(1-b)*Q(0)*Q(1) is an even number .

Hence contradiction.

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