## Saturday, February 11, 2006

### Diophantine Galore!. Topic: Number Theory. Level: AIME/Olympiad.

Problem #1: (360 Problems for Mathematical Contests - 2.1.19) Solve in the nonnegative integers the equation

$x^2+8y^2+6xy-3x-6y = 3$.

Solution: Write the equation as a quadratic in $x$, that is

$x^2+(6y-3)x+(8y^2-6y-3) = 0$.

The discriminant of this quadratic is $(6y-3)^2-4(8y^2-6y-3) = (2y-3)^2+12$. Since we are only looking for integral solutions, $(2y-3)^2+12$ has to be the square of a rational, but since it is integral, it must be the square of an integer. So $(2y-3)^2+12 = a^2$, but the difference between two squares must be the sum of consecutive odd integers and the only way $12$ can be written like that is $12 = 5+7$. So $2y-3 = 2$ (since the difference between $2^2$ and $3^2$ is $5$) is the only possible value, but that has no solutions. Hence our original equation has no solutions. QED.

--------------------

Problem #2: (360 Problems for Mathematical Contests - 2.1.24) Solve in the nonnegative integers the equation

$x+y+z+xyz = xy+yz+zx+2$.

Solution: Factor the given equation to get $(x-1)(y-1)(z-1) = 1$. So we have either $x-1 = y-1 = z-1 = 1$ or two of them negative and the other positive. This gives us the solutions

$(2,2,2); (2,0,0); (0,2,0); (0,0,2)$. QED.

--------------------

Comment: Two relatively simple diophantine equations (equations in which we are only looking for integral solutions), but they introduce important techniques. Rewriting as a quadratic and looking at the determinant is often a good way to go, especially if you can show the determinant is less than zero or something. Factoring is of course an extremely common method but a lot of the time the right way to factor may be difficult to find.

--------------------

Practice Problem: Try finding the factoring solution to the first problem (it exists).

#### 1 comment:

1. We factor the first equation into (x+4y-3)(x+2y)=3 ...but x+2y is nonnegative, so it must be 3 or 1. Casework from here and we get no solutions.