Q. \(y^2 = x^6 + 2x^3 + 4x^2 + 4x + 1\).

Problem: Find all integer solutions for the equation \(y^2 = x^6 + 2x^3 + 4x^2 + 4x + 1\)

1. Step 1 – Observe the structure of the equation:

   The right side of the equation is a polynomial of degree 6. Notice that the first two terms \(x^6 + 2x^3\) look like the beginning of the expansion of \((x^3 + 1)^2\). Let us expand that square:

   \((x^3 + 1)^2 = x^6 + 2x^3 + 1\).

2. Step 2 – Rewrite the original equation:

   We can rewrite the right side of the equation by substituting the square we just found:

   \(y^2 = (x^6 + 2x^3 + 1) + 4x^2 + 4x\)

   \(y^2 = (x^3 + 1)^2 + 4x(x + 1)\).

3. Step 3 – Use the bounding method for \(x > 0\) and \(x < -1\):

   For any integer \(x\) such that \(x > 0\) or \(x < -1\), the term \(4x(x + 1)\) is strictly positive. This means:

   \(y^2 > (x^3 + 1)^2\).

   Now consider the next consecutive perfect square, \((|x^3 + 1| + 1)^2\). For large values of \(|x|\), specifically \(|x| \ge 3\), the value of \(y^2\) will be trapped between two consecutive squares, \((x^3 + 1)^2\) and \((x^3 + 2)^2\), which is impossible for an integer \(y\). Therefore, we only need to check small integer values for \(x\).

4. Step 4 – Test specific integer values for \(x\):

   We test \(x\) values in the range \(-2, -1, 0, 1, 2\):

   • If \(x = -2\): \(y^2 = (-2)^6 + 2(-2)^3 + 4(-2)^2 + 4(-2) + 1 = 64 – 16 + 16 – 8 + 1 = 57\). 57 is not a perfect square.
   • If \(x = -1\): \(y^2 = (-1)^6 + 2(-1)^3 + 4(-1)^2 + 4(-1) + 1 = 1 – 2 + 4 – 4 + 1 = 0\). Thus, \(y = 0\).
   • If \(x = 0\): \(y^2 = 0^6 + 2(0)^3 + 4(0)^2 + 4(0) + 1 = 1\). Thus, \(y = 1\) or \(y = -1\).
   • If \(x = 1\): \(y^2 = 1^6 + 2(1)^3 + 4(1)^2 + 4(1) + 1 = 1 + 2 + 4 + 4 + 1 = 12\). 12 is not a perfect square.
   • If \(x = 2\): \(y^2 = 2^6 + 2(2)^3 + 4(2)^2 + 4(2) + 1 = 64 + 16 + 16 + 8 + 1 = 105\). 105 is not a perfect square.
5. Final answer:

   The integer solutions \((x, y)\) are \((-1, 0), (0, 1),\) and \((0, -1)\).

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