IMO 2007 Shortlist

Given that x >= y

Then gcd(x,y) =< y

So xy =< x + y +y = x + 2y

=> x*(y-1) - 2y + 2 =< 2

=> (x-2)(y-1) =<2 (1)

If x and y are both greater than or equal to 5

Hence (1) is incorrect. So x and y is less than 5

If x = 1, we have no solution

If x = 2, then y =3 or y = 4

If x = 3, then y = 3 or y =2

If x = 4, then y = 2

If we count (a,b) and (b,a) as 1, we have 3 solutions (4,2), (3,3), (3,2)

Let $gcd(x,y)=d$ .

Hence, $-x-y+xy=d$ which implies $1-x-y+xy=d+1$ or $(x-1)(y-1)=d+1$ .

So, $(x-1)|d$ and $(y-1)|d$

Hence, $x-1≤d$ and $y-1≤d$ .

Again, $d≤x$ and $d≤y$

So, $d≤x≤d+2$ and $d≤y≤d+2$

So $x=d,d+1$ or $d+2$ and $y=d,d+1$ or $d+2$ .

First,suppose $x=y$

Hence, $d=x$

So, $x^{2}=2x+x=3x$ .So, $x=3$ .Hence, $(x,y)=(3,3)$ is a solution.

The function $xy-x-y-gcd(x,y)$ is symmetric in $x,y$ .

From now on,we can assume $x<y$

If $x=d$ and $y=d+1$ ,then $d=gcd(x,y)=gcd(d,d+1)=1$

Hence, $x=1$ and $y=2$ .But this pair does not satisfy the given equation.

If $x=d$ and $y=d+2$ ,then $d(d+2)=d+(d+2)+d$ or $d^{2}-d-2=0$

Therefore, $(d+1)(d-2)=0$ .So, $d=2$ .

Hence, $(x,y)=(2,4)$

Finally,if $x=d+1$ and $y=d+2$ ,then $d=1$ .

Hence, $x=2$ and $y=3$ .See that this pair satisfies the given equation.

So, $(x,y)=(2,3)$ is another pair.

The equation is the same with the formula of the number of integer-coordinates point on the edges of a square triangle with edges x and y.

On the other hand, S = a + b/2 -1 (a and b is the number of the points inside and on the edges of the triangle, S is the area of the triangle.)

We have:

xy = x + y + gcd( x ; y )

S = xy/2

b = x + y +gcd( x ; y )

So xy/2 = a + ( x + y +gcd( x ; y ) ) / 2 - 1

Therefore a = 1, which means there's only 1 integer-coordinates point inside the triangle/

We can easily see that there are 3 solutions: (2;4) (2;3) and (3;3) Sorry for my poor English :P.