There are only 21 known non-cube $p

Question

I. Cuboids

The equation is connected to consecutive cubes equal to a cube like,

$$11^3+12^3+13^3+14^3=20^3$$

Given a cuboid with distinct sides $(a,b,c)=(p^2+q^2-1,\; 2p,\;2q)$,

$\hskip1.7in$

such that the sum of the squares of its sides is a square,

$$a^2+b^2+c^2 = (p^2+q^2-1)^2+(2p)^2+(2q)^2 = (p^2+q^2+1)^2$$

We seek those special cuboids such that the product of its sides divided by $2^5$ is a cube,

$$F(p,q)=\frac{abc}{2^5}=\frac{(p^2+q^2-1)(2p)(2q)}{2^5} = \frac{p\,q\,(p^2+q^2-1)}{2^3} = Y^3\tag1$$

which is nicely symmetric in $(p,q)$. Equivalently, it is a sum of $p$ consecutive cubes equal to the same cube,

$$F(p,q)=\sum_{k=0}^{p-1}\Big(\frac{1-p+q}2+k\Big)^3=Y^3\quad\tag2$$

Eq.1 and Eq.2 are just two sides of the same coin. For example, let $(p,q)=(3,8)$, then both yield $Y=6$,

$$F(3,8)=\sum_{k=0}^{3-1}\left(\tfrac{1-3+8}2+k\right)^3 = \sum_{k=0}^{2}\left(3+k\right)^3=3^3+4^3+5^3=6^3$$

Being lengths of a real cuboid, $(p,q)$ is required to be positive, and $\color{blue}{p<q}$ so the initial cube is also positive.

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II. Infinite family

There is an infinite family of solutions found in the 1800s when $p$ is a cube,

$$p = v^3,\quad q=\frac13(v^2 - 1)^2$$

though choose $v\neq 3m\,$ so $q\,$ is an integer.

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III. Sporadic solutions

A240970 gives a list of 49 values $p<10^5$. However, after removing 30 $p$ that are cubes, only 19 are left. But there are 2 more $p$ in the comments by Vladimir Pletser, one $p>10^5$ and the other $\color{blue}{p>10^7}$ (How did he find this?), so total of $19+2= 21$. About a third of the $p$ and $q$ are squares (Any underlying reason?).

The complete 21 $(p,q)$ are given below, counting the double solutions of $p=20,99,153$ only once,

$$\frac{p\,q\,(p^2+q^2-1)}{2^3} = Y^3$$

$p_k = 3, 4,\, 20,\, 20,\, 25,\, 49,\, 99,\; 99,\; 153,\; 153,\; 288,\; 1849,\; 6591,\; 10200,\, 13923, 14161, 19220, 21456, 25201, 33124, 49776, 63001, 176824, 11859210.$

$q_k = 8, 25, 25, 49, 36, 630, 120, 1210, 578, 1444, 833, 70304, 7200, 223109, 19942, 17408, 555025, 555025, 118580, 70225, 59535, 85340, 991875, 55278125.$

Note the largest is Pletser's $p=11859210=10\times33^4$. Also, the rather large square $q=555025 =745^2$ appears twice. All but five $p$'s were also found by Oleg567 in this old post.

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IV. Question

Sequence A240970 seems to be a brute force search for $p<10^5$. So how did Pletser later find in 2015 a pair with $p>10^5$ and $p>10^7$? Can we find more and does the method imply there are in fact infinitely many non-cube $p$?

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V. Updates

Note 1: Presumably an infinite family of elliptic curves is involved. A discussion in the old sci.math can be found here.

Note 2: Tomita found one $p\approx 10^6$ and Ezhov found four $q<10^7$ via exhaustive search, so the total now is $21+1+4=26$ solutions.

Note 3: It is preferred $(p,q)$ are positive integers with opposite parity, since initial cube $a_i=\frac12(1-p+q)$ must be an integer. So Tomita's $(p,q)=(6,6^2)$ and Ezhov's $(p,q)=(7200,92^2)$ in the comments yields half integers in the LHS,

$$\sum_{k=0}^{6-1} = \Big(\frac{31}2+k\Big)^3 = 33^3$$

$$\sum_{k=0}^{7200-1} = \Big(\frac{1265}2+k\Big)^3 = 97980^3$$

which are the only two solutions known so far.

Answer 1

Let $p=2^{n_1} \cdot 3^{n_2} \cdot 5^{n_3} \cdot 7^{n_4} \cdot 11^{n_5} \cdot 13^{n_6} \cdot 17^{n_7} \cdot 19^{n_8} ,$

$q=2^{m_1} \cdot 3^{m_2} \cdot 5^{m_3} \cdot 7^{m_4} \cdot 11^{m_5} \cdot 13^{m_6} \cdot 17^{m_7} \cdot 19^{m_8} $ where $0 \leq (n_1,n_2, \cdots, n_7,n_8)\leq 3 $ and $0 \leq (m_1,m_2, \cdots, m_7,m_8)\leq 3.$

Check $pq(p^2+q^2-1)$ is a cubic integer?

I found a solution $(p,q)=(2249390,8218925)=(2 \cdot 5 \cdot 11^3 \cdot 13^2, 5^2 \cdot 11^3 \cdot 13 \cdot 19).$

By increasing the prime factors, more solutions will be found.

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              p, factors
          [3, [3, 1]]
          [4, [2, 2]]
          [20, [2, 2], [5, 1]]
          [25, [5, 2]]
          [49, [7, 2]]
          [99, [3, 2], [11, 1]]
          [153, [3, 2], [17, 1]]
          [288, [2, 5], [3, 2]]
          [849, [3, 1], [283, 1]]
          [6591, [3, 1], [13, 3]]
          [10200, [2, 3], [3, 1], [5, 2], [17, 1]]
          [13923, [3, 2], [7, 1], [13, 1], [17, 1]]
          [14161, [7, 2], [17, 2]]
          [19220, [2, 2], [5, 1], [31, 2]]
          [21456, [2, 4], [3, 2], [149, 1]]
          [25201, [11, 1], [29, 1], [79, 1]]
          [33124, [2, 2], [7, 2], [13, 2]]
          [49776, [2, 4], [3, 1], [17, 1], [61, 1]]
          [63001, [251, 2]]
          [176824, [2, 3], [23, 1], [31, 2]]
          [11859210, [2, 1], [3, 4], [5, 1], [11, 4]]

Answer 2

Let $p\to 1/v$, then $v^2=((yv)^3/q - 1)/(q^2 - 1)$ and hyperellratpoints() from pari/gp can help find some $p$.

 my(H,v,p);
 parfor(q= 10^6, 10^7, if(!ispower(q,3),
  H= hyperellratpoints(('x^3/q - 1)/(q^2 - 1), 10^6);
  for(i= 1, #H,
   v= H[i][2];
   if(v>0, if(numerator(v)==1,
    p= 1/v;
    if(!ispower(p,3),
     print("("q","p")")
    )
   ))
  )
 ))

New $(p,q)$=(118580, 7216650), (351231, 1188208), (1012500, 1207296), (1997568, 4214809)