more random big primes
by math_explorer, Feb 13, 2011, 2:40 AM
APMO2001.4. A point in the plane with a cartesian coordinate system is called a mixed point if one of its coordinates is rational and the other one is irrational. Find all polynomials with real coefficients such that their graphs do not contain any mixed point.
First, suppose we have a polynomial of degree 2 or more whose graph doesn't contain any mixed point. Then its coefficients must all be rational, because we can express it as the sum of rational multiples of Lagrange polynomials by playing with the values we're given. Thus we can multiply the polynomial by the LCM of the coefficients' denominators to get an integer-coefficient polynomial whose graph also doesn't contain any mixed point.
Pick a huge prime
that is not a divisor of any of this polynomial's coefficients. Then if 
is an integer, the polynomial can't evaluate to 
when given a rational number (something with a denominator of 
is necessary). But since the polynomial has degree 2 or more, there must be numbers of the form in its range, so we have a mixed point.
So polynomials of degree 2 or more are not possible.
For polynomials of degree 1 or 0, it's obvious they work iff all of their coefficients are rational too.
APMO1996.2. Let
and be positive integers such that 
. Prove that
![\[ 2^n n! \leq \frac{(m+n)!}{(m-n)!} \leq (m^2 + m)^n \]](//latex.artofproblemsolving.com/5/0/b/50bf94bcc6c273eb32e8b0982003df811b70546d.png)
First, suppose we have a polynomial of degree 2 or more whose graph doesn't contain any mixed point. Then its coefficients must all be rational, because we can express it as the sum of rational multiples of Lagrange polynomials by playing with the values we're given. Thus we can multiply the polynomial by the LCM of the coefficients' denominators to get an integer-coefficient polynomial whose graph also doesn't contain any mixed point.
Pick a huge prime
that is not a divisor of any of this polynomial's coefficients. Then if 
is an integer, the polynomial can't evaluate to 
when given a rational number (something with a denominator of 
is necessary). But since the polynomial has degree 2 or more, there must be numbers of the form in its range, so we have a mixed point.So polynomials of degree 2 or more are not possible.
For polynomials of degree 1 or 0, it's obvious they work iff all of their coefficients are rational too.
APMO1996.2. Let
and be positive integers such that 
. Prove that![\[ 2^n n! \leq \frac{(m+n)!}{(m-n)!} \leq (m^2 + m)^n \]](http://latex.artofproblemsolving.com/5/0/b/50bf94bcc6c273eb32e8b0982003df811b70546d.png)
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