floor on polynomial

[Alpha Beta]

Let P,Q be non constant polynomials with real coefficients such that [P(x)]=[Q(x)] for all real x. ( [ ] denotes floor function). Show that P(x)=Q(x) for all real x.

[Nitin Prasad]

Let R(x)=P(x)-Q(x)

Now we have

$$[Q(x)]+[R(x)]\leq[P(x)]=[Q(x)+R(x)]\leq[Q(x)]+[R(x)]+1$$

Hence we have $$O\leq[R(x)]\leq1$$.

Now every non constant polynomial is unbounded. Hence degree of R(x) cannot be greater than 0.

Therefore degree of R(x) is zero and hence is a constant.

Now if P(x) is a nonconstant polynomial then there exist m such that P(m) is an integer and hence it can be shown that R(x)=O.

If degree of P(x) is 0 then the result follows trivially

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