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

Q: What are the integer solutions of y^2 = x^6 + 2x^3 + 4x^2 + 4x + 1?

The integer solutions are (x,y) = (0, ±1) and (−1, 0). These follow from factorization and bounding arguments (see next answers).

Q: How can I start solving y^2 = x^6 + 2x^3 + 4x^2 + 4x + 1?

Write the RHS as (x^3+1)^2 + 4x(x+1). Then set t = x^3+1 so (y−t)(y+t) = 4x(x+1). Analyze divisors and sizes to limit possibilities.

Q: Why use the substitution t = x^3 + 1?

It reduces the equation to y^2 − t^2 = 4x(x+1), i.e. (y−t)(y+t) = 4x(x+1). This factorization lets you compare divisor sizes and use gcd/parity to force small x-values.

Q: What gcd or parity arguments help eliminate other integer solutions?

Note gcd(y−t, y+t) divides 2; hence each factor is small up to sign. Combine that with 4x(x+1) factorization and size bounds to show only small x satisfy the equality, giving x = 0, −1.

Q: Can modular arithmetic quickly rule out many x?

Yes. Reduce the RHS modulo small primes (e.g. 3, 4, 8) to rule out residue classes that cannot be quadratic residues. This narrows possible x to tiny finite set to check directly.

Q: Is the polynomial on the right factorable over integers?

Not as simple product of low-degree integer polynomials. useful identity is RHS = (x^3+1)^2 + 4x(x+1), but no nontrivial integer factorization simplifies finding integer y.

Q: Are there rational (non-integer) solutions besides those from integer x?

The curve is hyperelliptic of genus 2, so only finitely many rational points. Determining all rational points requires advanced tools (descent, Chabauty); trivial rational points arise from integer solutions (0, ±1) and (−1, 0).

Q: What methods prove there are no large solutions?

Use size comparisons: for |x| large, (x^3+1)^2 dominates 4x(x+1), so |y−(x^3+1)| and |y+(x^3+1)| are tiny divisors of 4x(x+1). Those constraints force contradiction unless x is small; check remaining small x by direct computation.

Q: Where can I find full worked solution?

Look for solutions to "y^2 = x^6 + 2x^3 + 4x^2 + 4x + 1" on number theory/Diophantine equation resources, math competition solution repositories, or the Edubrain post referenced; they present the factorization-plus-divisor argument and case checks.

Answer

Compare to perfect squares.The expression is trapped between consecutive squares for large x.
\[ (x^3 + 1)^2 = x^6 + 2x^3 + 1 \]

\[ (x^3 + 2)^2 = x^6 + 4x^3 + 4 \]
Test small integer values.If x = 0, y^2 = 1, so y is 1 or -1.
If x = -1, y^2 = 0, so y is 0.
State the integer solutions.\[ (x, y) = (0, \pm 1), (-1, 0) \]

Detailed Explanation

Step-by-Step Solution for Integer Solutions of y² = x⁶ + 2x³ + 4x² + 4x + 1

To find all integer solutions (x, y) for this equation, we will use the method of bounding. This technique involves showing that for most values of x, the expression on the right side of the equation falls strictly between two consecutive perfect squares. If an integer y² is trapped strictly between n² and (n+1)² for some integer n, it is impossible for y to be an integer. By doing this, we can narrow down the possible values of x to a small set of numbers that we can test manually.

Step 1: Compare the expression to a known perfect square

First, we examine the leading terms of the polynomial: x⁶ + 2x³. This structure strongly resembles the expansion of the binomial (x³ + 1)². Let us calculate the exact expansion of this square:

(x³ + 1)² = x⁶ + 2x³ + 1

Now, we will subtract this perfect square from our original equation’s right-hand side to find the difference:

(x⁶ + 2x³ + 4x² + 4x + 1) – (x⁶ + 2x³ + 1) = 4x² + 4x

We can factor this difference as 4x(x + 1). We need to determine when this difference is positive. If x is any positive integer (meaning x ≥ 1), both x and x + 1 are positive, making the product 4x(x + 1) strictly greater than zero. Therefore, for all x ≥ 1, our expression is strictly greater than (x³ + 1)²:

y² > (x³ + 1)²

Step 2: Find an upper bound for positive values of x

Since we know our expression is larger than (x³ + 1)² for positive x, we will test the next consecutive perfect square, which is (x³ + 2)². Let us expand this square:

(x³ + 2)² = x⁶ + 4x³ + 4

We want to find out when this new square is larger than our expression. We subtract our original polynomial from this square:

(x⁶ + 4x³ + 4) – (x⁶ + 2x³ + 4x² + 4x + 1) = 2x³ – 4x² – 4x + 3

Let us define g(x) = 2x³ – 4x² – 4x + 3 and plug in small positive integers to see when it becomes positive:

For x = 1: g(1) = 2(1)³ – 4(1)² – 4(1) + 3 = 2 – 4 – 4 + 3 = -3 (The difference is negative)
For x = 2: g(2) = 2(8) – 4(4) – 4(2) + 3 = 16 – 16 – 8 + 3 = -5 (The difference is negative)
For x = 3: g(3) = 2(27) – 4(9) – 4(3) + 3 = 54 – 36 – 12 + 3 = 9 (The difference is positive)

For any x ≥ 3, the cubic term 2x³ will dominate the quadratic and linear terms, ensuring the difference remains positive. This means that for all integers x ≥ 3, our expression is strictly less than (x³ + 2)²:

y² < (x³ + 2)²

Step 3: Establish the bounds for negative values of x

We must also check negative integers. We return to the difference between our expression and the first square, which was 4x(x + 1). If x ≤ -2, both x and x + 1 are negative numbers. Multiplying two negative numbers yields a positive product. Thus, for all x ≤ -2:

y² > (x³ + 1)²

Because x is negative, x³ + 1 is actually larger than x³. So, the perfect square strictly below (x³ + 1)² is (x³)². Let us expand (x³)²:

(x³)² = x⁶

We now subtract our original expression from (x³)² to see if it provides an upper bound:

x⁶ – (x⁶ + 2x³ + 4x² + 4x + 1) = -2x³ – 4x² – 4x – 1

Let us test this difference for small negative integers (where x ≤ -2):

For x = -2: -2(-8) – 4(4) – 4(-2) – 1 = 16 – 16 + 8 – 1 = 7 (The difference is positive)
For x = -3: -2(-27) – 4(9) – 4(-3) – 1 = 54 – 36 + 12 – 1 = 29 (The difference is positive)

This pattern holds true for all x ≤ -2. Therefore, our expression is strictly less than (x³)²:

y² < (x³)²

Step 4: Analyze the bounded regions

From Step 2, we proved that for x ≥ 3:

(x³ + 1)² < y² < (x³ + 2)²

From Step 3, we proved that for x ≤ -2:

(x³ + 1)² < y² < (x³)²

In both of these domains, y² is strictly wedged between the squares of two consecutive integers. It is mathematically impossible for a perfect square to exist between two consecutive perfect squares. Therefore, we have proven that no integer solutions can exist for x ≥ 3 or for x ≤ -2.

Step 5: Manually test the remaining integer candidates

The only integers that are not covered by our bounding arguments are x = -1, x = 0, x = 1, and x = 2. We will substitute each of these directly into the original equation y² = x⁶ + 2x³ + 4x² + 4x + 1 to check for perfect squares.

Case x = -1:

y² = (-1)⁶ + 2(-1)³ + 4(-1)² + 4(-1) + 1

y² = 1 – 2 + 4 – 4 + 1

y² = 0

The square root of 0 is an integer (y = 0). This gives us the coordinate pair (-1, 0).

Case x = 0:

y² = (0)⁶ + 2(0)³ + 4(0)² + 4(0) + 1

y² = 0 + 0 + 0 + 0 + 1

y² = 1

Taking the square root yields two integer solutions: y = 1 and y = -1. This gives us the coordinate pairs (0, 1) and (0, -1).

Case x = 1:

y² = (1)⁶ + 2(1)³ + 4(1)² + 4(1) + 1

y² = 1 + 2 + 4 + 4 + 1

y² = 12

Since 12 is not a perfect square, there is no integer solution for y when x = 1.

Case x = 2:

y² = (2)⁶ + 2(2)³ + 4(2)² + 4(2) + 1

y² = 64 + 16 + 16 + 8 + 1

y² = 105

Since 105 is not a perfect square, there is no integer solution for y when x = 2.

Final Answer

By evaluating the bounded limits and testing the remaining domain, the complete set of integer solutions (x, y) for the given equation is:

(x, y) ∈ { (-1, 0), (0, 1), (0, -1) }

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FAQ

What are the integer solutions of y^2 = x^6 + 2x^3 + 4x^2 + 4x + 1?

The integer solutions are (x,y) = (0, ±1) and (−1, 0). These follow from factorization and bounding arguments (see next answers).

How can I start solving y^2 = x^6 + 2x^3 + 4x^2 + 4x + 1?

Write the RHS as (x^3+1)^2 + 4x(x+1). Then set t = x^3+1 so (y−t)(y+t) = 4x(x+1). Analyze divisors and sizes to limit possibilities.

Why use the substitution t = x^3 + 1?

It reduces the equation to y^2 − t^2 = 4x(x+1), i.e. (y−t)(y+t) = 4x(x+1). This factorization lets you compare divisor sizes and use gcd/parity to force small x-values.

What gcd or parity arguments help eliminate other integer solutions?

Note gcd(y−t, y+t) divides 2; hence each factor is small up to sign. Combine that with 4x(x+1) factorization and size bounds to show only small x satisfy the equality, giving x = 0, −1.

Can modular arithmetic quickly rule out many x?