Polynomials :{}{}{}{}{}{}}

f(x) = an xⁿ + an-1 xⁿ⁻¹ + ... + a1 x + a0
a)f(r/s) = 0
an (r/s)ⁿ + an-1 (r/s)ⁿ⁻¹ + ... + a1 (r/s) + a0 = 0
an rⁿ + an-1 srⁿ⁻¹ + ... + a1 sⁿ⁻¹ r + sⁿ a0 = 0 ...... (*)
rⁿ = - ( an-1 srⁿ⁻¹ + ... + a1 sⁿ⁻¹ r + sⁿ a0 ) / an- ( an-1 srⁿ⁻¹ + ... + a1 sⁿ⁻¹ r + sⁿ a0 ) is even since each term have an even factor 's' ,
by given that an is odd therefore rⁿ = even / odd = even , and hence r is even.

b)If s is even , by the result of part (a) , r is also even , then s and r have a common factor 2. (Contradiction)
So s must be odd.By (*) ,
an rⁿ + an-1 srⁿ⁻¹ + ... + a1 sⁿ⁻¹ r + sⁿ a0 = 0an rⁿ + [an-1 (s-1)rⁿ⁻¹ + an-1 rⁿ⁻¹] + ... + [a1 (sⁿ⁻¹ - 1) r + a1 r] + [a0 (sⁿ-1) + a0] = 0 [an rⁿ + an-1 rⁿ⁻¹ + ... + a1 r + a0] + [an-1 (s-1)rⁿ⁻¹ + ... + a1 (sⁿ⁻¹ - 1) r + a0 (sⁿ-1)] = 0f(r) = - [an-1 (s-1)rⁿ⁻¹ + ... + a1 (sⁿ⁻¹ - 1) r + a0 (sⁿ-1)]
- [an-1 (s-1)rⁿ⁻¹ + ... + a1 (sⁿ⁻¹ - 1) r + a0 (sⁿ-1)] is even since each term have an even factor 'sᵏ - 1' since s is odd were 1 ≤ k ≤ n is an integer. Hence f(r) is even.

c)Let r/s be a rational root of f(x) = 0 and r and s are relatively prime ,
by the result of b) , f(r) = an rⁿ + an-1 rⁿ⁻¹ + ... + a1 r + a0 is even , ==> an rⁿ + an-1 rⁿ⁻¹ + ... + a1 r = even - a0 = even - odd = odd==> r (an rⁿ⁻¹ + an-1 rⁿ⁻² + ... + a1) = odd==> r must be odd Then start from an rⁿ + an-1 rⁿ⁻¹ + ... + a1 r + a0 = even ==>
[an (rⁿ-1) + an-1 (rⁿ⁻¹-1) + ... + a1 (r-1)] + [an + an-1 + ... + a1 + a0] = even==>
an + an-1 + ... + a1 + a0 = even - [an (rⁿ-1) + an-1 (rⁿ⁻¹-1) + ... + a1 (r-1)] ==>
f(1) = even - even = even
since each term of [an (rⁿ-1) + an-1 (rⁿ⁻¹-1) + ... + a1 (r-1)] have an even factor
'rᵏ - 1' since r is odd were 1 ≤ k ≤ n is an integer.