Solution June 28, 2007
Table of contents |
Problem 1
Find all positive integers such that and are all prime numbers.
Solution 1
Remember that 2 is the only even prime. We consider two cases.
- n even: Then 3n − 4 is an even prime so that 3n − 4 = 2 and n = 2. The resulting numbers 2,3,7 are indeed prime so this is a solution.
- n odd: Then 5n − 3 is an even prime, and 5n − 3 = 2. This implies n = 1 which isn't a solution because 3 − 4 = − 1 is not a prime number.
So n = 2 is the only solution to the problem.
Problem 2
and are prime numbers. has distinct rational roots. Find all and which work.
Solution 2
We can write x^{2} − px + q = (x − a)(x − b) where a and b are the roots of the polynomial.
By the rational roots theorem (http://en.wikipedia.org/wiki/Rational_root_theorem) (if this is too scary, see the note below), the roots of x^{2} − px + 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 x^{2} − px + q are
They are only rational when p^{2} − 4q is a perfect square. If p is even then p^{2} − 4q is even and so is its square root. Similarily, if p is odd then p^{2} − 4q is odd and its square root is odd. In either case, are even, so the roots are integers if they are rational.