# On the elliptic curves $u \big(u + (n - 1)^2\big) \big(u + (n + 1)^2\big) = y_1^2$ and $v(v - 1)(v - n^2) = y_2^2$

A certain problem in equal sums of like powers for $7$th powers entails the elliptic curve,

$$u(u+127^2)(u+129^2) = y^2$$

I was looking at the general case,

$$u \big(u + (n - 1)^2\big) \big(u + (n + 1)^2\big) = y_1^2\tag1$$

and using this online Magma to test integer $n>1$ such that it has positive rank. It starts as,

$$n = 6, 13, 16, 18, 22, 23, 32, 33, 35, 36, 37, 41, 42, 43, 44, 45, 46, 50,\dots$$

Checking the OEIS, it turns out it may be sequence A228380 but involves the $n$ of the elliptic curve,

$$v(v - 1)(v - n^2) = y_2^2\tag2$$

such that it has positive rank.

Q: Is it true that the sequence of $n>1$ for $(1)$ and $(2)$ are in fact the same? If so, why?

Let $E_n$ and $F_n$ be the elliptic curves \begin{aligned} E_n : y_1^2 &= u\bigl(u+(n-1)^2\bigr)\bigl(u+(n+1)^2\bigr), \\ F_n : y_2^2 &= v(v-1)(v-n^2). \end{aligned} There are only finitely many $n$ values for which $E_n$ and $F_n$ are isomorphic, since $$\text{numerator of }j(E_n)-j(F_n) = (n + 3)(3 n + 1)(n^2 - 2 n + 5)(5 n^2 - 2 n + 1)(n^3 - 3 n^2 - n - 1)(n^3 + n^2 + 3 n - 1).$$ So if the ranks are matching, best guess is that the curves are 2-isogenous. Thus using the formulas in Example 4.5 of The Arithmetic of Elliptic Curves, we have $$F_n : y^2=x^3+ax^2+bx\quad\text{with a=-(n^2+1) and b=n^2.}$$ The 2-isogenous curve $F_n/\langle(0,0)\rangle$ is $$E : Y^2 = X^3-2aX^2+rX\quad\text{with r=a^2-4b=(n^2-1)^2.}$$ Hence the isogenous curve is $$E : Y^2=X^3+2(n^2+1)X^2+(n^2-1)^2X = X(X+(n-1)^2)(X+(n+1)^2),$$ which is exactly your curve $E_n$. In particular, since the curves are $2$-isogenous over $\mathbb Q$, we have $\text{rank }E_n(\mathbb Q)=\text{rank }F_n(\mathbb Q)$. This holds for all $n\in\mathbb Q\setminus\{\pm1,0\}$.
The MO question On the elliptic curve $x(x+a^2)(x+b^2) = y^2$ is related.