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Problem of the Week
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.

A closed curve forming a shape like a circle that has been
horizontally compressed and then rotated.

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 \(15\).

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 \((13,59)\).