A Family of Mordell Equations

Relevant wiki: General Diophantine Equations - Problem Solving

We have $y^2=(x-p) \big(x^2+xp+p^2\big). \qquad (1)$ Let $d=\gcd \big(x-p, x^2+xp+p^2\big)$ , then $d \mid x^2+xp+p^2-(x-p)^2=3xp \ \ \ \ \Longrightarrow \ \ \ \ d \mid 3xp$ Suppose that $d>1$ . Then $d$ cannot take $3$ or $p$ , because it leads to $d \mid x-p \mid y$ and negates conditions of the problem. It follows that $d \mid x$ and $d \mid x-(x-p)=p$ , and again contradicts $p \nmid y$ . Thus $d=1$ and factors of the RHS of equation $(1)$ are co-prime and consequently both of them are squares. Specially $x^2+xp+p^2=n^2$ . $\Delta_x$ must be a perfect square, that is $-3p^2+4n^2=k^2$

• If $p=2$ , then $(2n-k)(2n+k)=12$ . Thus, $k=2l$ and $(n-l)(n+l)=3$ . It gives $n=2$ and $x^2+2x+4=4$ . But both $x=0$ and $x=-2$ contradict $x>p$ .

• If $p=3$ , then $(2n-k)(2n+k)=27$ . It gives $n=3, 7$ . For $n=3$ we get $x^3+3x=0$ and no integer solution for $y$ . For $n=7$ we get $x^3+3x=40$ , which gives $x=-8$ and $x=5$ . Its clear that $x=-8$ is unacceptable and for $x=5$ , $x-p=2$ is not a perfect square.

Henceforth $p>3$ and $(2n-k)(2n+k)=3p^2$ Since divisors of $3p^2$ are $1<3<p<3p<p^2<3p^2$ , then we have $k=p$ , $k=(p^2-3)/2$ or $k=(3p^2-1)/2$ .

• For $k=p$ we get $x-p=(-k+p)/2-p=-p$ , which is not a perfect square.

• For $k=(p^2-3)/2$ , $x-p$ becomes $(p^2-6p-3)/4$ . Thus $p^2-6p-3=m^2$ and $m^2+12=(p-3)^2$ is a perfect square, which is possible only when $m=2$ . It gives $p=7$ and the solutions $(p, x, y)=\color{#D61F06}(7, 8, \pm13)$ .

• If $k=(3p^2-1)/2$ , then $x-p=(3p^2-6p-1)/4$ . Thus $3p^2-6p-1=m^2$ and $\Delta_p=3m^2+12=a^2$ . Therefore $a=3b$ and $m^2+4=3b^2$ . Since a perfect square is congruent to $0, 1$ mod $3$ , then $m^2+4$ cannot be divisible by $3$ .Thus there is no solution in this case.

Thus, with the given conditions, there are two integer points on the family of Mordell curves $y^2=x^3-p^3$ . They are $(p, x, y)=\color{#D61F06} (7, 8, \pm13)$ . So the answer is $\boxed{30}$ .

Working in the Euclidean domain $\mathbb{Z}[\omega]$ we are trying to solve $y^2 \; = \; (x - p)(x - \omega p)(x - \omega^2p) \hspace{2cm} x,y \in \mathbb{Z}$ Since neither $3$ nor $p$ divides $y$ , no element of $\mathbb{Z}[\omega]$ with norm $3$ or $p$ can divide $y$ . Any irreducible common factor of $x-p$ and $x - \omega p$ would have to divide $(1-\omega)p$ and $y$ ; this is not possible (since $1-\omega$ has norm $3$ ). Similarly, $x-p$ and $x - \omega^2p$ cannot have a common factor. Nor can $x - \omega p$ and $x - \omega^2p$ have a common factor. Thus the three numbers $x-p,x-\omega p,x-\omega^2p$ are pairwise coprime. Thus we can write $x - p \; = \; uA^2 \hspace{1cm} x - \omega p = vB^2 \hspace{1cm} x - \omega^2p = wC^2$ where $A,B,C \in \mathbb{Z}[\omega]$ and $u,v,w$ are units in $\mathbb{Z}[\omega]$ .Since the units in $\mathbb{Z}[\omega]$ are either squares or negative squares, we can assume that $u,v,w = \pm1$ .

If $A = m + n\omega$ then $A^2 = (m^2 - n^2) + n(2m-n)\omega$ . Since $A^2 \in \mathbb{R}$ we deduce that $n(2m-n) = 0$ , so either $n=0$ and $A^2 = m^2$ or else $n = 2m$ and $A^2= -3m^2$ . Since $3$ does not divide $y$ , we deduce that the first of these cases must be true, and so $x - p = \pm m^2$ for some $m \in \mathbb{N}$ . Since $(x - p\omega)(x - p\omega^2) = x^2 + xp + p^2$ is positive, we deduce that $x-p = m^2$ .

We now consider the fact that $x - p\omega \; = \; \varepsilon B^2 \; = \; \varepsilon(b + c\omega)^2 \; = \; \varepsilon\big(b^2 - c^2 + c(2b - c)\omega\big)$ where $b,c$ are integers and $\varepsilon = \pm1$ , For definiteness, assume that $c > 0$ . Then we have $x - p \; = \; m^2 \hspace{2cm} x \; = \; \varepsilon(b^2 - c^2) \hspace{2cm} -p \; = \; \varepsilon c(2b-c)$ There are four cases to consider:

• If $\varepsilon= 1\,,\,c = 1$ then $b = \tfrac12(1-p)$ (so that $p \ge 3$ is odd). Thus $x = b^2 - c^2 = \tfrac14(p^2 - 2p - 3)$ , so that $\begin{aligned} m^2 & = x - p \; = \; \tfrac14(p^2 - 6p - 3) \; = \; \tfrac14\big((p-3)^2 - 12\big) \\ 12 & = (p-3)^2 - 4m^2 \; = \; (p-3 - 2m)(p-3+2m) \end{aligned}$ Since $p-3-2m$ and $p-3+2m$ must both be positive with the same parity, it follows that $p-3-2m=2$ , $p-3+2m=6$ , so that $p=7$ , $m=1$ , $x=8$ and so $y^2 = 169$ , and hence $y = \pm13$ .
• If $\varepsilon = 1\,,\,c=p$ , then $b = \tfrac12(p-1)$ (so that $p$ is again odd). Thus $x = b^2 - c^2 = \tfrac14(1 - 2p - 3p^2)$ , so that $m^2 = x-p = \tfrac14(1 - 6p - 3p^2) < 0$ , which is impossible.
• If $\varepsilon = -1\,,\,c=1$ , then $b = \tfrac12(p+1)$ (so that $p$ is again odd). Then $x = c^2- b^2 = \tfrac14(3 - 2p - p^2)$ , so that $m^2 = x-p = \tfrac14(3 - 6p - p^2) < 0$ , which is impossible.
• If $\varepsilon = -1\,,\,c=p$ , then $b = \tfrac12(p+1)$ (so that $p$ is again odd). Then $x = c^2 - b^2 = \tfrac14(3p^2 - 2p - 1)$ , so that $\begin{aligned} m^2 & = x - p \; =\; \tfrac14(3p^2 - 6p - 1) \; = \; \tfrac14\big(3(p-1)^2 - 4\big) \\ 4(m^2 + 1) & = 3(p-1)^2 \end{aligned}$ which is impossible, since $4(m^2+1)$ cannot be divisible by $3$ .

Thus the only solutions to this family of equations are $x=8$ , $y = \pm13$ , $p = 7$ , making the answer $(8+7) + (8+7) = \boxed{30}$ .