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.
1this one can be done by contradiction,
let the real root of f(x)=a
=>f(x)= (x-a)(g(x))
=> f(0).f(1) = (a)(a-1)(blah-blah)
=> RHS is always even as in the pair of (a,a-1) one has to be even while the other has to be odd.(we are discussing over integer roots)
=> this is a contradiction as by the fiven condition L.H.S = odd.odd = odd.
thus our assumption is wrong and f(x) = 0 has no integer solutions.
21I want to use the fact that for a polynomial over Z, (a-b)| (f(a)- f(b))
Let f(t) = 0
t is an integer
t|f(t)- f(0) or t|f(0)
t-1|f(t) - f(1) or t-1| f(1)
f(0) is odd
t must be odd
then t-1 is even, if t-1|f(1) , then f(1) must also be even,which is a contradiction...
I know it's verbatim Rishabh's :P
