Solution June 28, 2007

Table of contents

Problem 1

Find all positive integers n\, such that 3n-4,4n-5,\, and 5n-3\, are all prime numbers.

Solution 1

Remember that 2 is the only even prime. We consider two cases.

So n = 2 is the only solution to the problem.

Problem 2

p\, and q\, are prime numbers. x^2\, -\,p\,x\, +\,q\  =\  0 has distinct rational roots. Find all p\, and q\, which work.

Solution 2

We can write x2px + q = (xa)(xb) where a and b are the roots of the polynomial.

By the rational roots theorem ( (if this is too scary, see the note below), the roots of x2px + q = 0 are integers that divide q.

The product of the two roots must be q so we have a = − 1,b = − q or a = 1,b = q.

The sum of the two roots is equal to p which rules out the first possibility, because p > 0.

Now we have p = q + 1 where p and q are both prime. The only possibility for that is p = 3 and q = 2. That answers the problem.

Note: Instead of using the rational roots theorem, we can have a closer look at the quadratic formula. The roots of x2px + q are

x_{1,2} = \frac{p\pm\sqrt{p^2-4q}}2

They are only rational when p2 − 4q is a perfect square. If p is even then p2 − 4q is even and so is its square root. Similarily, if p is odd then p2 − 4q is odd and its square root is odd. In either case, p\pm\sqrt{p^2-4q} are even, so the roots are integers if they are rational.