Solve in $\mathbb{N}$ the equation $3x^2 + 2x = y^2$.

Question

Solve in $\mathbb{N}$ the equation $3x^2 + 2x = y^2$.

After factoring, we get $x(3x+2)=y^2$. If LHS is a perfect square and we let $x = n^2z$, where $z$ is not divisible by the square of any prime number, and $3x+2 = zk^2$ and I got stuck here. Any help would be appreciated.

Answer 1

This can be reduced to a Pell's equation.

$$x(3x +2) = y^2$$

$$(x,3x+2) = (x,2)$$

Assume $x$ to be even. Then the equation is reduced to,

$$k(3k+1) = (\frac{y}{2})^2$$

Here as $(k,3k+1) = (k,1)$, $k = a^2$ and $3k+1 = b^2$

Pell's equation. $$3a^2 + 1 = b^2$$

$x = 2a^2$ and $y = 2ab$

Edit:

Pell's equation has infinite solutions, this can be shown.

$a= 1$ and $b = 2$ is a pair solution.

Observe: $$(2+3^{\frac{1}{2}})(2-3^{\frac{1}{2}}) = 1$$

$${((2+3^{\frac{1}{2}})(2-3^{\frac{1}{2}})})^n = 1^n$$

By binomial expansion.

$$(2+3^{\frac{1}{2}})^n = x_n +y_n3^{\frac{1}{2}}$$ $$(2-3^{\frac{1}{2}})^n = x_n -y_n3^{\frac{1}{2}}$$

$${((2+3^{\frac{1}{2}})(2-3^{\frac{1}{2}})})^n = 1^n = x_n^2-3y_n^2$$

For every $n$ there is a solution,

$n=2$

$$(2+3^{\frac{1}{2}})^2 = 7+4.(3)^{\frac{1}{2}}$$ $$x_n = b = 7,y_n = a = 4$$

This works also.

If odd then, $(x,2)=1$

$x = z^2$

$3x +2 = f^2$

Which can not happen as we are getting $f^2 \equiv 2 \pmod{3}$

Answer 2

Multiply by $3$ and set $z:=3x$ to find that $$3y^2=9x^2+6x=z^2+2z=(z+1)^2-1,$$ and hence for $X:=z+1$ and $Y:=y$ we get the Pell equation $$X^2-3Y^2=1.\tag{1}$$ This equation is well studied, see Wikipeda for example. Every solution $(X_n,Y_n)$ to $(1)$ is of the form $$X_n+Y_n\sqrt{3}=(2+\sqrt{3})^n,$$ for some integer $n$. You get a solution to the original equation whenever $X\equiv1\pmod{3}$, which is whenever $n$ is even.