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While investigating a system involving "equal sums of like powers", an elliptic curve popped up,

$$9 (1 + 4 n^4)^2 + 30 (4 + n^2 - 24 n^4 + 4 n^6) x^2 + 5 (32 - 40 n^2 + 53 n^4) x^4 = y^2$$

Some easy rational points are,

$$x = (0,\; 1,\; n)$$

Using the tangent-chord method, from the last two, we get,

$$\; x = \frac{6 (1 + 4 n^4) (17 + 12 n^2)}{151 - 49 n^2 + 144 n^4 + 144 n^6}$$ $$x = \frac{\; 6 (1 + 4 n^4) (3n + 23 n^3)}{9 + 9 n^2 - 79 n^4 + 121 n^6}$$

Question: But are there rational points where the numerator and denominator are polynomials of degree less than 6, preferably only quadratics?

Tito Piezas III
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1 Answers1

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Looking at a bunch of examples, it appears that your curve (which is isomorphic to an elliptic curve) has rank $2$ over $\mathbb{Q}(n)$. Using this, one can find some other rational points of degree $\leq 6$, but not where the numerator or denominator is a quadratic. In particular, there are solutions when $$ x = \frac{12 n^{4} + 3}{12 n^{3} \pm 23n^{2} + 17n \pm 3}. $$

Jeremy Rouse
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  • Thanks. Substituting this into the system, I end up with another curve $F(n) = z^2$ where $F(n)$ is of degree 7. Sigh. (I was hoping it would be at most only of degree 4 thus isomorphic to an elliptic curve. I will post the system tomorrow.) – Tito Piezas III Feb 11 '23 at 13:15
  • This MO post is the context to the curve above. Essentially, the curve is the discriminant of the quartic condition in the MO post (with some variables specialized). – Tito Piezas III Feb 15 '23 at 06:23
  • Given the form $ax^4+bx^2+c^2 = y^2$, can the point you found be expressed in terms of general $a,b,c,$ or is it specific only to this curve? – Tito Piezas III Feb 19 '23 at 02:20
  • No - the points I found are specific to the coefficients $a$, $b$ and $c$ you specified in the question. I would guess that for a generic $a$, $b$ and $c$, the corresponding elliptic curve has rank $1$ over $\mathbb{Q}(a,b,c)$. – Jeremy Rouse Feb 19 '23 at 02:58
  • Ah ok, thanks. I'm using various approaches to solve Choudhry's system in the MO post, but I keep getting $F(n) = z^2$ where $F(n)$ has degree $>4$. Sigh. – Tito Piezas III Feb 19 '23 at 03:07
  • Still no headway in Choudhry's system, but you may be interested in this elliptic curve. – Tito Piezas III Feb 20 '23 at 04:04