## January 2015 solution

Curiously enough, functions satisfying all of the criteria not only exist, but are common and far from esoteric.

Consider the function f(x)=x3-5x.

The function is clearly continuous, because it is a polynomial.

It is clearly a surjection, because it is a continuous function that ranges from -∞ to ∞, like any odd-degree polynomial.

It is not an injection, because from -sqrt(5/3) to sqrt(5/3) its derivative is negative. (Elsewhere it is positive.) This means that all f(x) values in the range between ± 10 sqrt(15)/9 have three separate x values leading to them.

Because it is a polynomial, it is also closed in ℚ.

It is not a surjection when restricted to ℚ because, for example, it can never take the value 1/2. To see this, suppose that some p/q exists such that f(p/q)=1/2. This would indicate that p3/q3-5p/q = (p3-5pq2)/q3 = 1/2. However, for this to be true, q must be even. Assuming p/q is given in reduced form, this indicates that p is odd. However, this would mean that p3-5pq2 is odd, while q3 is divisible by 8, so in reduced form this fraction must have a denominator that is a multiple of 8. In particular, it cannot equal 1/2.

It is most interesting, however, to note that g is an injection. To show this, let us assume to the contrary that two rationals exist that map to the same value.

We will refer to these two values as x=p1/q and y=p2/q, having assumed that we have already brought them to a common denominator. (The forms presented here may therefore not be fully reduced as individual rational numbers. However, we will assume that they are reduced jointly, in the sense that we are using the minimum possible q value.)

First off, note that if x3-5x=y3-5y, then

x3-y3=5x-5y

(x2+xy+y2)(x-y)=5(x-y)

x2+xy+y2=5,

where the last step follows from xy.

Multiplying both sides by q2, we get that

p12+p1p2+p22 ≡ 0 (mod 5),
which has as its only solution that both p1 and p2 are divisible by 5. Unfortunately, when that is the case, the value of p12+p1p2+p22 divides by 25, so q must also divide by 5.

This would indicate that the pair (p1/q, p2/q), as a pair, is not in reduced form, as all three integers involved could have been divided by 5. This contradicts our minimality assumption.

The function is therefore an injection over ℚ.

Q.E.D.