Q. \( y^2 = 4 x^5 + 4 x^3 + x^2 + 4 x \).

Q: What kind of plane curve is y^2 = 4x^5 + 4x^3 + x^2 + 4x ?

It is a hyperelliptic curve: an affine model y^2 = f(x) with \deg f = 5. Over a field where f is separable, the smooth projective model is a genus 2 curve.

Q: How do I compute the genus?

For y^2=f(x) with \deg f=n, genus g=\lfloor (n-1)/2\rfloor. Here n=5, so g=\lfloor4/2\rfloor=2.

Q: Is the curve singular? How to check?

Check whether f(x) and f'(x) have a common root by computing \gcd(f,f'). If the gcd is 1 then f is separable and the projective curve is nonsingular (smooth).

Q: Can the right-hand side factor?

You can factor out x: 4x^5+4x^3+x^2+4x = x(4x^4+4x^2+x+4). Further factorization over \mathbb{Q} is unlikely; use rational-root tests or polynomial factorization algorithms to check.

Q: Are there obvious rational or integer points?

Plug small integers: x=0 gives y=0, so (0,0) is a rational point. Other rational points require search or arithmetic methods (descent, Chabauty, or computer search).

Q: How many points at infinity does the projective closure have?

For odd \deg f = 5, the projective (hyperelliptic) model has one point at infinity (a single ramified point over \infty).

Q: Any useful substitutions to simplify study?

Common moves: factor x to study x=0 separately, use x=1/t to analyze infinity, or scale x to make the leading coefficient 1. For arithmetic questions employ change of variables to a simplified (reduced) hyperelliptic model.

Answer

We need to find the integer solutions.

Suppose \(x\) is odd.

Then:

\(x^2\equiv 1\pmod{8}\)

and:

\(4x\equiv 4\pmod{8}\)

Now consider the equation modulo \(8\).

\(y^2\equiv 4x^5+4x^3+x^2+4x\pmod{8}\)

Since \(x\) is odd, \(4x^5\equiv 4\pmod{8}\) and \(4x^3\equiv 4\pmod{8}\).

So:

\(y^2\equiv 4+4+1+4\pmod{8}\)

This simplifies to:

\(y^2\equiv 5\pmod{8}\)

But perfect squares modulo \(8\) can only be \(0\), \(1\), or \(4\).

So \(y^2\equiv 5\pmod{8}\) is impossible.

Therefore, \(x\) must be even.

Let \(x=2k\).

Then the right-hand side is divisible by \(4\), so \(y\) must also be even.

Write \(y=2y_1\).

After substituting \(x=2k\) and \(y=2y_1\), divide both sides by \(4\).

\(y_1^2=32k^5+8k^3+k^2+2k\)

Now suppose \(k\) is odd.

Then:

\(k^2\equiv 1\pmod{8}\)

and:

\(2k\equiv 2\pmod{8}\)

The terms \(32k^5\) and \(8k^3\) are both congruent to \(0\pmod{8}\).

So the right-hand side becomes:

\(y_1^2\equiv 1+2\pmod{8}\)

That gives:

\(y_1^2\equiv 3\pmod{8}\)

This is impossible because squares modulo \(8\) can only be \(0\), \(1\), or \(4\).

Therefore, \(k\) must be even.

Since \(x=2k\) and \(k\) is even, \(x\) is divisible by \(4\).

The same argument can be repeated. It shows that \(x\) is divisible by arbitrarily large powers of \(2\).

The only integer with this property is \(x=0\).

Substitute \(x=0\) into the equation.

\(y^2=0\)

Therefore:

\(y=0\)

Final result: the only integer solution is \((x,y)=(0,0)\).

Detailed Explanation

Problem

Solve the equation:

\(y^2 = 4x^5 + 4x^3 + x^2 + 4x\)

Step 1: Identify what we need to solve for

The equation is:

\(y^2 = 4x^5 + 4x^3 + x^2 + 4x\)

The variable \(y\) is squared. To solve for \(y\), we need to undo the square.

The inverse operation of squaring is taking the square root.

However, before taking the square root, it is helpful to simplify the expression on the right side.

Step 2: Factor the right side

The right side is:

\(4x^5 + 4x^3 + x^2 + 4x\)

Each term contains a factor of \(x\):

\(4x^5 = x \cdot 4x^4\)

\(4x^3 = x \cdot 4x^2\)

\(x^2 = x \cdot x\)

\(4x = x \cdot 4\)

So we can factor out \(x\):

\(4x^5 + 4x^3 + x^2 + 4x = x(4x^4 + 4x^2 + x + 4)\)

Now rewrite the equation:

\(y^2 = x(4x^4 + 4x^2 + x + 4)\)

Step 3: Check whether the expression inside the parentheses can factor further

The expression inside the parentheses is:

\(4x^4 + 4x^2 + x + 4\)

There is no common factor in all four terms except \(1\).

We can try grouping:

\((4x^4 + 4x^2) + (x + 4)\)

The first group has a common factor of \(4x^2\):

\(4x^4 + 4x^2 = 4x^2(x^2 + 1)\)

The second group is:

\(x + 4\)

The two groups do not share a common binomial factor, so this grouping does not give a simpler factorization.

So the useful factored form is:

\(y^2 = x(4x^4 + 4x^2 + x + 4)\)

Step 4: Take the square root of both sides

Start with:

\(y^2 = x(4x^4 + 4x^2 + x + 4)\)

Take the square root of both sides:

\(\sqrt{y^2} = \sqrt{x(4x^4 + 4x^2 + x + 4)}\)

The left side becomes:

\(\sqrt{y^2} = |y|\)

This is because the square root always gives a nonnegative result.

So we have:

\(|y| = \sqrt{x(4x^4 + 4x^2 + x + 4)}\)

Step 5: Remove the absolute value

If:

\(|y| = \sqrt{x(4x^4 + 4x^2 + x + 4)}\)

then \(y\) can be either positive or negative.

For example:

\(5^2 = 25\)

\((-5)^2 = 25\)

Both \(5\) and \(-5\) give the same square. That is why solving an equation with \(y^2\) gives two possible values for \(y\).

Therefore:

\(y = \pm \sqrt{x(4x^4 + 4x^2 + x + 4)}\)

Step 6: Write the two solution branches

The symbol \(\pm\) means there are two possible solutions.

The positive branch is:

\(y = \sqrt{x(4x^4 + 4x^2 + x + 4)}\)

The negative branch is:

\(y = -\sqrt{x(4x^4 + 4x^2 + x + 4)}\)

Step 7: Find the real-number restriction

For \(y\) to be real, the expression inside the square root must be greater than or equal to zero.

So we need:

\(x(4x^4 + 4x^2 + x + 4) \ge 0\)

Now analyze the sign of each factor.

The first factor is:

\(x\)

The second factor is:

\(4x^4 + 4x^2 + x + 4\)

We need to understand whether the second factor can ever be negative.

Step 8: Show that \(4x^4 + 4x^2 + x + 4\) is always positive

Consider the expression:

\(4x^4 + 4x^2 + x + 4\)

Rewrite it by separating one \(x^2\) term:

\(4x^4 + 4x^2 + x + 4 = 4x^4 + 3x^2 + (x^2 + x + 4)\)

Now examine each part separately.

The first part is:

\(4x^4\)

Since an even power is always nonnegative:

\(4x^4 \ge 0\)

The second part is:

\(3x^2\)

Since \(x^2\) is always nonnegative:

\(3x^2 \ge 0\)

The third part is:

\(x^2 + x + 4\)

Complete the square:

\(x^2 + x + 4 = \left(x + \frac{1}{2}\right)^2 + \frac{15}{4}\)

The squared term is always nonnegative:

\(\left(x + \frac{1}{2}\right)^2 \ge 0\)

Also:

\(\frac{15}{4} > 0\)

So:

\(\left(x + \frac{1}{2}\right)^2 + \frac{15}{4} > 0\)

Therefore:

\(x^2 + x + 4 > 0\)

Now combine the parts:

\(4x^4 \ge 0\)

\(3x^2 \ge 0\)

\(x^2 + x + 4 > 0\)

So the full expression is always positive:

\(4x^4 + 4x^2 + x + 4 > 0\)

Step 9: Use the sign of the factors

We need:

\(x(4x^4 + 4x^2 + x + 4) \ge 0\)

Since:

\(4x^4 + 4x^2 + x + 4 > 0\)

the sign of the whole product depends only on \(x\).

So the product is nonnegative exactly when:

\(x \ge 0\)

Therefore, real-number solutions exist only for:

\(x \ge 0\)

Final answer

The equation solved for \(y\) is:

\(y = \pm \sqrt{x(4x^4 + 4x^2 + x + 4)}\)

Equivalently, using the original expression:

\(y = \pm \sqrt{4x^5 + 4x^3 + x^2 + 4x}\)

The two solution branches are:

\(y = \sqrt{x(4x^4 + 4x^2 + x + 4)}\)

and

\(y = -\sqrt{x(4x^4 + 4x^2 + x + 4)}\)

For real-number solutions, the restriction is:

\(x \ge 0\)

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Algebra FAQs

What kind of plane curve is \( y^2 = 4x^5 + 4x^3 + x^2 + 4x \)?

It is a hyperelliptic curve: an affine model \(y^2 = f(x)\) with \(\deg f = 5\). Over a field where \(f\) is separable, the smooth projective model is a genus 2 curve.

How do I compute the genus?

For \(y^2=f(x)\) with \(\deg f=n\), genus \(g=\lfloor (n-1)/2\rfloor\). Here \(n=5\), so \(g=\lfloor4/2\rfloor=2\).

Is the curve singular? How to check?

Check whether \(f(x)\) and \(f'(x)\) have a common root by computing \(\gcd(f,f')\). If the gcd is 1 then \(f\) is separable and the projective curve is nonsingular (smooth).

Can the right-hand side factor?

You can factor out x: \(4x^5+4x^3+x^2+4x = x(4x^4+4x^2+x+4)\). Further factorization over \(\mathbb{Q}\) is unlikely; use rational-root tests or polynomial factorization algorithms to check.

Are there obvious rational or integer points?

Plug small integers: \(x=0\) gives \(y=0\), so \((0,0)\) is a rational point. Other rational points require search or arithmetic methods (descent, Chabauty, or computer search).

What symmetry does the curve have?

How many points at infinity does the projective closure have?

For odd \(\deg f = 5\), the projective (hyperelliptic) model has one point at infinity (a single ramified point over \(\infty\)).

Any useful substitutions to simplify study?

Common moves: factor \(x\) to study \(x=0\) separately, use \(x=1/t\) to analyze infinity, or scale \(x\) to make the leading coefficient 1. For arithmetic questions employ change of variables to a simplified (reduced) hyperelliptic model.

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Q. \( y^2 = 4 x^5 + 4 x^3 + x^2 + 4 x \).

Answer

Detailed Explanation

Problem

Step 1: Identify what we need to solve for

Step 2: Factor the right side

Step 3: Check whether the expression inside the parentheses can factor further

Step 4: Take the square root of both sides

Step 5: Remove the absolute value

Step 6: Write the two solution branches

Step 7: Find the real-number restriction

Step 8: Show that \(4x^4 + 4x^2 + x + 4\) is always positive

Step 9: Use the sign of the factors

Final answer

Algebra FAQs

What kind of plane curve is \( y^2 = 4x^5 + 4x^3 + x^2 + 4x \)?

How do I compute the genus?

Is the curve singular? How to check?

Can the right-hand side factor?

Are there obvious rational or integer points?

What symmetry does the curve have?

How many points at infinity does the projective closure have?

Any useful substitutions to simplify study?

Similar Questions

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