Problem E and Solution

Points on an Ellipse

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 \((x,y)=(5,10)\).

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 \((13,59)\).