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

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

1. Step 1 – Compare the expression to a known perfect square:

   Notice that the first two terms \(x^6 + 2x^3\) are part of the expansion of \((x^3 + 1)^2\). We can expand this square to compare:
   \[
   (x^3 + 1)^2 = x^6 + 2x^3 + 1.
   \]
   The difference between the given curve and this square is:
   \[
   (x^6 + 2x^3 + 4x^2 + 4x + 1) – (x^6 + 2x^3 + 1) = 4x^2 + 4x = 4x(x + 1).
   \]

2. Step 2 – Identify intervals where the expression lies between squares:

   If \(x>0\) or \(x<-1\), the product \(4x(x + 1)\) is strictly positive. This means for these values: \[ y^2 > (x^3 + 1)^2.
   \]
   Now compare the expression to the next consecutive square, \((x^3 + 2)^2\):
   \[
   (x^3 + 2)^2 = x^6 + 4x^3 + 4.
   \]
   Subtracting our expression from this gives:
   \[
   (x^6 + 4x^3 + 4) – (x^6 + 2x^3 + 4x^2 + 4x + 1) = 2x^3 – 4x^2 – 4x + 3.
   \]
   For \(x\ge 3\), this difference is positive, meaning
   \[
   y^2 < (x^3 + 2)^2. \] When an integer \(y^2\) is strictly between two consecutive squares, it cannot be a perfect square itself.

3. Step 3 – Test the remaining integer candidates:

   Because the bounding logic excludes \(x\ge 3\) and \(x\le -2\), we only need to manually check the integers \(x = -1, 0, 1, 2\).

4. Step 4 – Evaluate \(x = -1\):

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

   \(y^2 = 1 – 2 + 4 – 4 + 1 = 0\).

   Thus \(y = 0\). This gives the solution \((-1, 0)\).

5. Step 5 – Evaluate \(x = 0\):

   \(y^2 = 0^6 + 2\cdot0^3 + 4\cdot0^2 + 4\cdot0 + 1 = 1\).

   Thus \(y = \pm 1\). This gives the solutions \((0, 1)\) and \((0, -1)\).

6. Step 6 – Evaluate \(x = 1\) and \(x = 2\):

   For \(x = 1\): \(y^2 = 1 + 2 + 4 + 4 + 1 = 12\) (not a square).

   For \(x = 2\): \(y^2 = 64 + 16 + 16 + 8 + 1 = 105\) (not a square).

7. Final Answer:

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