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Problem:- Let $$\text{P(x)}=x^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\dots+a_{1}x+a_{0}$$ be a polynomial with integer coefficients,such that,$$\text{P(0) and P(1)}$$  are odd integers.Show that

(a)$$\text{P(x)}$$ does not have any even integer roots.

(b)$$\text{P(x)}$$ does not have any odd integer roots.

Solution:- Given the two statements (a) and (b)  above it is clear that if we can prove that $$\text{P(x)}$$ has no integer roots,then we are done.

Proof:- Let us assume $$\text{P(x)}$$ has an integer root $$\text{ a}$$.

Then we can write,

$$\text{P(x)=(x-a)Q(x)}\dots(1)$$

where,$$\text{Q}$$is any function of $$\text{ x}$$.

Now ,putting $$\text{x=0,1}$$ in $$\text{ 1}$$,we get

$$\text{P(0)=(-a)Q(0)}\dots(2)$$

and,

$$\text{P(1)=(1-a)Q(1)}\dots(3)$$

now as $$\text{(1-a)}$$ and $$\ \text{(-a)}$$,are consecutive integers $$\text{(2)}$$ and $$\text{(3)}$$

cannot be both odd,

which means that $$\text{P(0),P(1)}$$ cannot be both odd,given whatever $$\text{Q(0) and Q(1)}$$ are.

So,we can conclude that there exists no integer solution of $$\text{P(x)}$$.