Q. \(y^2 + y = x^3 – x^2 – 10x – 20\) elliptic curve.

Analysis of the Elliptic Curve \(y^2 + y = x^3 – x^2 – 10x – 20\)

To find the integer solutions for the elliptic curve defined by the equation \(y^2 + y = x^3 – x^2 – 10x – 20\), we follow a structured algebraic approach.

1. Step 1 – Complete the square for the y-terms:

   The left side of the equation is \(y^2 + y\). To complete the square, we multiply the entire equation by 4 to avoid working with fractions immediately:

   \[4(y^2 + y) = 4(x^3 – x^2 – 10x – 20)\]

   \[4y^2 + 4y = 4x^3 – 4x^2 – 40x – 80\]

   Now, add 1 to both sides to complete the square on the left:

   \[4y^2 + 4y + 1 = 4x^3 – 4x^2 – 40x – 80 + 1\]

   \[(2y + 1)^2 = 4x^3 – 4x^2 – 40x – 79\]

2. Step 2 – Analyze the condition for integer solutions:

   For \(x\) and \(y\) to be integers, the right-hand side \(4x^3 – 4x^2 – 40x – 79\) must be a perfect square. Let \(D = 4x^3 – 4x^2 – 40x – 79\). We need to find integer values of \(x\) such that \(D = k^2\) for some integer \(k\).

3. Step 3 – Test integer values for x:

   Testing small integer values for \(x\):

   • If \(x = 5\): \[D = 4(5)^3 – 4(5)^2 – 40(5) – 79 = 4(125) – 4(25) – 200 – 79 = 500 – 100 – 200 – 79 = 121.\]
   • Since \(121 = 11^2\), this is a perfect square.

   Checking other small integers (like \(x = 4, 3, 2, 1, 0\) or higher values) does not yield other perfect squares. For instance, if \(x < 5\), the value of \(D\) becomes negative or fails to be a square.

4. Step 4 – Solve for y:

   Using the result from Step 3 where \((2y + 1)^2 = 11^2\):

   \(2y + 1 = 11\) or \(2y + 1 = -11\)

   Solving the first case: \(2y = 10\), so \(y = 5\).

   Solving the second case: \(2y = -12\), so \(y = -6\).

5. Final Answer:

   The integer points on the elliptic curve are \((5, 5)\) and \((5, -6)\).

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