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Let the polynomial be

Notice that f(0)=a_0  and f(1)=a_0+a_1+a_2+\cdots+a_n

If c is a root of f(x), then f(c)=0

If c=2k is even :
f(2k)=a_0+2(stuff)=\text{odd+even}=\text{odd}, which can not be 0.

If c=2k+1 is odd :
f(2k+1)=a_0+a_1+a_2+\cdots+a_n+2(stuff)=\text{odd+even}=\text{odd}, which can not be 0.

f(c) is an odd integer in both cases and consequently there are no integer roots.
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