On the quartic $\sum_{k=1}^m (a+k)^4 = (a+1)^4+(a+2)^4+ \dots +(a+m)^4 = z^4$ and elliptic curves?

Question

I. Fourth powers

I've asked about the equation,

$$\sum_{k=1}^m (a+k)^4 = z^4$$

in a post ten years ago and there may be some progress. The closest I've come is 34 fourth powers almost equal to a fourth power,

$$\sum_{k=1}^{34} (4+k)^4 = 5^4+6^4+\dots+38^4 = 64^4+\color{red}{19^4}$$

Note: I find it strange that the "extra" happens to be a fourth power and is half the largest term on the LHS side. But if we relax conditions a bit and allow the initial term $a$ to be a quadratic irrational $a+b\sqrt{d}$, then there is a solution with $(2n+1)^2=15^2=225$ addends,

$$\sum_{k=1}^{225} \left(-113+\sqrt{\frac{164}{15}}+k\right)^4=292^4$$

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II. Elliptic curve

It turns out one needs to solve a rather simple elliptic curve in $x$,

$$30 (p - x)(p - 3 p^2 + 2 n p + 30 p x - 15 x^2)=y^2$$

with parameter,

$$p = 2n(n + 1) (2 n + 1)(2 n^2 + 2n +1)$$

However, the only $n$ I've found so far with non-trivial solution $(x\neq p)$ is when $n=7$ of the above. The method is to expand,

$$\sum_{k=1}^{\alpha} \left(-\beta+\sqrt{v}+k\right)^4=Av^2+Bv+C$$

where $(A,B,C)$ are integers if,

\begin{align} \alpha &= (2n+1)^2\\[4pt] \beta &= \,n^2+(n+1)^2 = 2n^2+2n+1 \end{align}

The RHS is a fourth power,

$$Av^2+Bv+C = z^4$$

if we define $v$ as,

$$v = \frac{p - 3 p^2 + 2 p n + 15 x^2}{30 (1 + 2 n) (p - x)}$$

using the rational solution $x$ to the elliptic curve above.

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III. Example

Let $n=7$, so $\alpha=15^2=225$, and $\beta = 113$, then expanding,

$$\sum_{k=1}^{225} \left(-113+\sqrt{v}+k\right)^4=15 (480510352 + 379680 v + 15 v^2)$$

The elliptic curve is,

$$30(- 189840 + x) (108114829200 - 5695200 x + 15 x^2)=y^2$$

with non-trivial solution $x=85100$. The formula for $v$ yields $v=\dfrac{164}{15}$ so,

$$15 (480510352 + 379680 v + 15 v^2) = 292^4$$

Therefore,

$$\sum_{k=1}^{225} \left(-113+\sqrt{\frac{164}{15}}+k\right)^4=z^4=292^4$$

Using this fixed parameter $n=7$ and the initial rational point $x$, we can find infinitely more $(v,z)$ that uses $225$ addends.

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IV. Question

Given the elliptic curve,

$$30 (p - x)(p - 3 p^2 + 2 n p + 30 p x - 15 x^2)=y^2$$

with parameter,

$$p = 2n(n + 1) (2 n + 1)(2 n^2 + 2n +1)$$

for what $n\neq7$ can we find a non-trivial solution $(x\neq p)$ ?

Answer 1

$$y^2=30 (p - x)(p - 3 p^2 + 2 n p + 30 p x - 15 x^2) \tag{1}$$ with $$p = 2n(n+1)(2n+1)(2n^2+2n+1)$$

An equation $(1)$ can be transformed to en elliptic curve $E.$

$$E: Y^2 = 450X^3+(-1860n^2-300000n^6-115200n^9-345600n^7-259200n^8-66600n^4-23040n^{10}-60n-15600n^3-174240n^5)X$$ with $$x = (-1860n^2-300000n^6-115200n^9-345600n^7-259200n^8-66600n^4-23040n^{10}-60n-15600n^3-174240n^5)/X+8n^5+20n^4+20n^3+10n^2+2n$$

Finding the generators of $E$, we got the solutions where $n<100.$

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$n$ with rank $0$ are $11,30,35,39,42,49,51,52,54,56,59,63,69,70,74,75,77,78,80,83,89,96,99.$

[n,x]
[2, 146705/32]
[3, 11270/9]
[4, 341579742840/90193009]
[6, 3004216407108390868849/67738239469559808]
[7, 465108]
[8, 182337399834/190969]
[9, 25664620785/132098]
[10, 219625219605275198/48936076225]
[13, 13970525418006606/3991586041]
[14, 254021235956678844074493091678162476/90542051319188791807597465369]
[16, 19295408]
[17, 10241721949365948019423958940/7226157970284039274489]
[21, 4681364008090/516961]
[22, 138062122489/1568]
[25, 1479985649545/4418]
[28, 1954315061358284870/635997961]
[45, 74748129502046/25281]
[53, 171611404889036070/21836929]
[55, 13261476883352258764225608662160/189851439732240640201]
[66, 71686619774/9]

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(Edit by Piezas). Thanks to Tomita's work, we can find more equalities. For example, using the values in the comments:

For $n=2$ with $x=4\times163=652$, so $\alpha=(2n+1)^2=25$,

$$\sum_{k=1}^{25} \left(-13+\sqrt{\frac{75921}{320}}+k\right)^4=\sum_{k=1}^{25} \left(-13+\sqrt{- \frac{175761}{320}}+k\right)^4=\left(\frac72\right)^{12}$$

for $n=3$ with $x=8988$, so $\alpha=(2n+1)^2=49$,

$$\sum_{k=1}^{49} \left(-25+\sqrt{-\frac{427}9}+k\right)^4=\sum_{k=1}^{49} \left(-25+\sqrt{-\frac{10373}9}+k\right)^4=\left(\frac{91}3\right)^{4}$$

for $n=7$ with $x=85100$, so $\alpha=(2n+1)^2=225$,

$$\sum_{k=1}^{225} \left(-113+\sqrt{\frac{164}{15}}+k\right)^4=\sum_{k=1}^{225} \left(-113+\sqrt{-\frac{379844}{15}}+k\right)^4=(292)^{4}$$

and infinitely more for each solvable $n$ (and they come in pairs).