# Problem of the Week Problem E and Solution Points on an Ellipse

## Problem

The graph of $$(x+1)^2+(y-2)^2=100$$ is a circle with centre $$(-1,2)$$ and radius $$10$$.

The graph of $$10x^2-6xy + 4x +y^2=621$$ is shown below. The shape of this curve is known as an ellipse.

List all the ordered pairs $$(x,y)$$ of non-negative integers $$x$$ and $$y$$ that satisfy the equation $$10x^2-6xy + 4x +y^2=621$$.

Note: When solving this problem, it might be useful to use the following idea.

By completing the square, $x^2 + y^2 + 2x - 4y= 95$ can be rewritten as $(x+1)^2 + (y-2)^2=100$ One solution to this equation is $$(x,y)=(5,10)$$.

## Solution

Starting with the given equation, we obtain the following equivalent equations: \begin{aligned} 10x^2-6xy + 4x +y^2 &= 621\\ 9x^2 - 6xy +y^2 + x^2 + 4x &= 621\\ 9x^2 - 6xy +y^2 + x^2 + 4x + 4 &= 621 + 4\\ (3x-y)^2 + (x+2)^2 &= 625 \end{aligned}

Notice that $$625 = 25^2$$.
Since $$x$$ and $$y$$ are both integers, then the left side of the given equation is the sum of two perfect squares. Since any perfect square is non-negative, then each of these perfect squares is at most $$625=25^2$$.

The pairs of perfect squares that sum to 625 are $$625$$ and $$0$$, $$576$$ and $$49$$, and $$400$$ and $$225$$.

Therefore, $$(3x-y)^2$$ and $$(x+2)^2$$ are equal to $$25^2$$ and $$0^2$$ in some order, or $$24^2$$ and $$7^2$$ in some order, or $$20^2$$ and $$15^2$$ in some order.

Furthermore, $$3x-y$$ and $$x+2$$ are equal to $$\pm 25$$ and $$\pm 0$$ in some order, or $$\pm 24$$ and $$\pm 7$$ in some order, or $$\pm 20$$ and $$\pm 15$$ in some order.

Since $$x \geq 0$$, then $$x+2 \geq 2$$. So we need to consider when $$x+2$$ is equal to $$25$$, $$24$$, $$7$$, $$20$$, or $$15$$.

• If $$x+2 = 25$$, then $$x=23$$. Also, $$3x-y=0$$. Thus, $$y=69$$. Since $$x\geq 0$$ and $$y\geq 0$$, $$(23, 69)$$ is a valid ordered pair.

• If $$x+2 = 24$$, then $$x=22$$. Also, $$3x-y=7$$ or $$3x-y=-7$$.
When $$3x-y=7$$, we find $$y=59$$. Since $$x\geq 0$$ and $$y\geq 0$$, $$(22, 59)$$ is a valid ordered pair.
When $$3x-y=-7$$, we find $$y=73$$. Since $$x\geq 0$$ and $$y\geq 0$$, $$(22, 73)$$ is a valid ordered pair.

• If $$x+2 = 7$$, then $$x=5$$. Also, $$3x-y=24$$ or $$3x-y=-24$$.
When $$3x-y=24$$, we find $$y=-9$$. Since $$y< 0$$, this does not lead to a valid ordered pair.
When $$3x-y=-24$$, we find $$y=39$$. Since $$x\geq 0$$ and $$y\geq 0$$, $$(5, 39)$$ is a valid ordered pair.

• If $$x+2 = 20$$, then $$x=18$$. Also, $$3x-y=15$$ or $$3x-y=-15$$.
When $$3x-y=15$$, we find $$y=39$$. Since $$x\geq 0$$ and $$y\geq 0$$, $$(18, 39)$$ is a valid ordered pair.
When $$3x-y=-15$$, we find $$y=69$$. Since $$x\geq 0$$ and $$y\geq 0$$, $$(18, 69)$$ is a valid ordered pair.

• If $$x+2 = 15$$, then $$x=13$$. Also, $$3x-y=20$$ or $$3x-y=-20$$.
When $$3x-y=20$$, we find $$y=19$$. Since $$x\geq 0$$ and $$y\geq 0$$, $$(13, 19)$$ is a valid ordered pair.
When $$3x-y=-20$$, we find $$y=59$$. Since $$x\geq 0$$ and $$y\geq 0$$, $$(13, 59)$$ is a valid ordered pair.

Therefore, the ordered pairs of non-negative integers that satisfy the equation are $$(23,69)$$, $$(22,59)$$, $$(22,73)$$, $$(5,39)$$, $$(18,39)$$, $$(18,69)$$, $$(13,19)$$, and $$(13,59)$$.