Solving the diophantine equation x^3=y^2+2

This will be a post illustrating how unique factorization in the ring \mathbb{Z}[\sqrt{-2}]=\{a+b\sqrt{-2}: a, b\in\mathbb{Z}\} helps us solve the diophantine equation x^3=y^2+2. We first start by proving

Lemma: The ring \mathbb{Z}[\sqrt{-2}] is a UFD.

Proof. Define N(a+b\sqrt{-2})=a^2+2b^2. The identity

(a^2+2b^2)(c^2+2d^2)=(ac-2bd)^2+2(ad+bc)^2

proves that

N(a+b\sqrt{-2})N(c+d\sqrt{-2})=N((a+b\sqrt{-2})(c+d\sqrt{-2}))

So N is a multiplicative function. We claim that N is a Euclidean norm on \mathbb{Z}[\sqrt{-2}]. Let \alpha, \beta\in \mathbb{Z}[\sqrt{-2}] with \beta\neq 0. We need to show that there exist elements q, r\in \mathbb{Z}[\sqrt{-2}] such that \alpha=q\beta + r where N(r)<N(\beta). Now \alpha/\beta\in\mathbb{Q}[\sqrt{-2}], so we can write \alpha/\beta=x+y\sqrt{-2} where x, y\in\mathbb{Q}. Let m, n be the closest integers to x and y, respectively. Let q=m+n\sqrt{-2}\in\mathbb{Z}[\sqrt{-2}]. Since |x-m|\leq 1/2 and |y-n|\leq 1/2, it follows that

N(\alpha/\beta - q)= N((x-m)+(y-n)\sqrt{-2})=(x-m)^2+2(y-n)^2\leq \dfrac{3}{4}

Set r=\alpha-q\beta \in\mathbb{Z}[\sqrt{-2}]. Then \alpha=q\beta+r and

\displaystyle N(r)=N(\alpha-q\beta)=N\left(\frac{\alpha}{\beta}-q\right) \cdot N(\beta) \leq \frac{3}{4} N(\beta)<N(\beta)

as desired.

Let’s turn our attention in solving the equation x^3=y^2+2 in integers x and y. Suppose x, y\in\mathbb{Z} satisfy x^3=y^2+2. First, x cannot be even because then y^2+2\equiv (\text{mod } 4) which is impossible. So x must be odd, and as a result y is also odd. Now we have the factorization x^3=(y+\sqrt{-2})(y-\sqrt{-2}) in \mathbb{Z}[\sqrt{-2}]. We first show that y+\sqrt{-2} and y-\sqrt{-2} are relatively prime. Suppose \gamma\in\mathbb{Z}[\sqrt{-2}] divides both y+\sqrt{-2} and y-\sqrt{-2}. Then \gamma divides their difference 2\sqrt{-2}=-(\sqrt{-2})^3. If \gamma is not a unit, then \sqrt{-2} would divide \gamma. This is because \sqrt{-2} is prime: indeed, if \sqrt{-2}=\alpha\beta, then from N(\sqrt{-2})=2, we see that N(\alpha)=1 or N(\beta)=1; so either \alpha or \beta is a unit; i.e \sqrt{-2} is irreducible and hence prime because we are in a UFD. But if \sqrt{-2} divides \gamma, then \sqrt{-2} divides y+\sqrt{-2} and so \sqrt{-2} divides y. But this means y=\sqrt{-2}(k+r\sqrt{-2})=-2r + k\sqrt{-2}, i.e. k=0 and y=-2r. But then y would be even, contradiction. We conclude that \gamma is a unit.

So y+\sqrt{-2} and y-\sqrt{-2} are relatively prime. So if p is any prime that divides x, then p^3 divides x^3 = (y+\sqrt{-2})(y-\sqrt{-2}). So p^3 divides either y+\sqrt{-2} or y-\sqrt{-2}. Hence, y+\sqrt{-2} must be of the form u h^3 for some h\in\mathbb{Z}[\sqrt{-2}] where u is a unit. Since the only units in \mathbb{Z}[\sqrt{-2}] are \pm 1 (units correspond to N(u)=1) which are already cubes, we can absorb them into h^3 term. So write h=a+b\sqrt{-2} for some a, b\in\mathbb{Z}. Then we get

y+\sqrt{-2} = (a+b\sqrt{-2})^3 = a^3-6ab^2+(3a^2b-2b^3)\sqrt{-2}

As a result, y=a(a^2-6b^2) and 1=b(3a^2-2b^2). We deduce that b=\pm 1. This gives 3a^2-2=\pm 1. Clearly 3a^2-2=-1 doesn’t give any solution in integers; so 3a^2-2=1 and consequently, a=\pm 1. So y=a(a^2-6b^2)=\pm 5. Now x^{3}=y^2+2 gives x=3. Therefore, (x, y)=(3, \pm 5) are the only solutions in integers.

 

 

 

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