Show that if a polynomial with integer coefficients has a rational root, then the rational root must be an integer. Hence, show that x^n-5x+7=0 has no rational roots.

Let f(x)=x^n+a_(n-1)x^(n-1)+...+a_0 with n>=2 and a_i an integer have a rational root x=p/q.

Consider q^(n-1)f(p/q).

q^(n-1)f(p/q)=p^n/q+a_(n-1)p^(n-1)+...+a_0q^(n-1)=0

==> a_(n-1)p^(n-1)+...+a_0q^(n-1)=-p^n/q

The LHS of this equation is a sum of integers. Thus it is also an integer. Thus the RHS must be an integer, so p/q is an integer.

Consider x^n-5x+7=0.

We now know that is this has a rational root, then it has an integral root. But if it has a root in the integers, it must have a root modulo 2.

Consider x^n-5x+7=0 mod 2.

If x=0 mod 2, then x^n-5x+7=0-0+1=1 mod 2. If x=1 mod 2, then x^n-5x+7=1-1+1=1 mod 2.

Hence, either way there are no solutions as the LHS is always odd and the RHS is even.

Thus x^n-5x+7=0 has no rational solutions.