I am trying to find all integer solutions $(x,y)$ to the following equation:
$$ \frac{x^3-x}{6} + 1 = y^2 \quad \cdots\ (\ast) $$
(Note: By setting $n=x-1$, this problem is equivalent to finding all integers $n$ for which the $n$-th tetrahedral number plus one is a perfect square. My motivation for this question came from observing that for the current year 2025, we have $2024+1=2025=45^2$, where 2024 is the 22nd tetrahedral number. This corresponds to the solution $(x,y)=(23,45)$. A related entry can be found in the On-Line Encyclopedia of Integer Sequences as A216268.)
This is a problem of finding all integer points on an elliptic curve.
First, note that if $(x_0, y_0)$ is a solution, then $(x_0, -y_0)$ is also a solution. Therefore, we can assume $y \geq 0$ without loss of generality. Also, since the right-hand side must be non-negative, we have $(x^3-x)/6 + 1 \geq 0$, which simplifies to $x^3-x+6 \geq 0$, or $(x+2)(x^2-2x+3) \geq 0$. Since $x^2-2x+3 = (x-1)^2+2 > 0$ for all real $x$, this inequality requires $x \geq -2$.
A numerical search for solutions in the range $-2 \leq x \leq 10^8$ yields the following 10 solutions for $y \geq 0$:
$$ \begin{align*} (x, y) = & (-2, 0), \\ & (-1, 1), \\ & (0, 1), \\ & (1, 1), \\ & (6, 6), \\ & (9, 11), \\ & (23, 45), \\ & (25, 51), \\ & (64, 209), \\ & (13776, 660099) \end{align*} $$
I would like to prove that this is the complete set of solutions for $y \geq 0$.
My Attempt
We can rearrange $(\ast)$ into:
$$ (x+2)(x^2-2x+3) = 6y^2. $$
Let $x' = x + 2$. The equation becomes:
$$ x'(x'^2 - 6x' + 11) = 6y^2 \quad \cdots \ (\ast\ast) $$
Let $z' = x'^2 - 6x' + 11$. We know $z' = (x'-3)^2+2 \geq 2$. Since $x \geq -2$, we have $x' = x+2 \geq 0$.
The case $x'=0$ gives the solution $(x, y) = (-2, 0)$. From now on, we will only consider the case where $x' > 0$ is a positive integer.
Next, we find the greatest common divisor of $x'$ and $z'$:
$$ \begin{align*} \gcd(x', z') &= \gcd(x', x'^2 - 6x' + 11) \\ &= \gcd(x', 11) \\ &=\begin{cases} 1 & \text{if } 11 \nmid x' \\ 11 & \text{if } 11 \mid x' \end{cases} \end{align*} $$
This leads to two main cases.
Case 1: $\gcd(x', z') = 1$ (i.e., $11 \nmid x'$ )
Since $x'$ and $z'$ are coprime, we can analyze the equation $x'z' = 6y^2$ by distributing the factors. This gives four possibilities for some positive integers $u, v$:
$$ \begin{array}{llll} \text{(a)} & x' = 6u^2, & z' = v^2, & \text{with } \gcd(v, 6) = 1 \\ \text{(b)} & x' = 3u^2, & z' = 2v^2, & \text{with } \gcd(u, 2) = 1, \gcd(v, 3) = 1 \\ \text{(c)} & x' = 2u^2, & z' = 3v^2, & \text{with } \gcd(u, 3) = 1, \gcd(v, 2) = 1 \\ \text{(d)} & x' = u^2, & z' = 6v^2, & \text{with } \gcd(u, 6) = 1 \\ \end{array} $$
Case 2: $\gcd(x', z') = 11$ (i.e., $11 \mid x'$ )
If $11 \mid x'$, it follows from $x'z' = 6y^2$ that $11 \mid y$. Let $x' = 11x''$ and $y = 11y''$ for some positive integers $x'', y''$. Substituting these into $(\ast\ast)$ gives:
$$ x''(11x''^2 - 6x'' + 1) = 6y''^2 $$
In this new equation, $\gcd(x'', 11x''^2 - 6x'' + 1) = \gcd(x'', 1) = 1$. So we can apply the same logic as in Case 1, leading to four more possibilities:
$$ \begin{array}{llll} \text{(e)} & x'' = 6u^2, & 11x''^2 - 6x'' + 1 = v^2, & \text{with } \gcd(v, 6) = 1 \\ \text{(f)} & x'' = 3u^2, & 11x''^2 - 6x'' + 1 = 2v^2, & \text{with } \gcd(u, 2) = 1, \gcd(v, 3) = 1 \\ \text{(g)} & x'' = 2u^2, & 11x''^2 - 6x'' + 1 = 3v^2, & \text{with } \gcd(u, 3) = 1, \gcd(v, 2) = 1 \\ \text{(h)} & x'' = u^2, & 11x''^2 - 6x'' + 1 = 6v^2, & \text{with } \gcd(u, 6) = 1 \\ \end{array} $$
This reduces the problem to solving the following eight lemmas. I am using $X$ to represent either $x'$ or $x''$.
Lemma a:
The system $X=6u^2, X^2-6X+11 = v^2$ has no positive integer solutions $(X, u, v)$.
Lemma b:
The only positive integer solutions $(X, u, v)$ to the system $X=3u^2, X^2-6X+11 = 2v^2$ are $(3, 1, 1)$ and $(27, 3, 17)$.
Lemma c:
The only positive integer solutions $(X, u, v)$ to the system $X=2u^2, X^2-6X+11 = 3v^2$ are $(2, 1, 1)$, $(8, 2, 3)$, and $(13778, 83, 7953)$.
Lemma d:
The only positive integer solutions $(X, u, v)$ to the system $X=u^2, X^2-6X+11 = 6v^2$ are $(1, 1, 1)$ and $(25, 5, 9)$.
Lemma e:
The only positive integer solution $(X, u, v)$ to the system $X=6u^2, 11X^2-6X+1 = v^2$ is $(6, 1, 19)$.
Lemma f:
The system $X=3u^2, 11X^2-6X+1 = 2v^2$ has no positive integer solutions $(X, u, v)$.
Lemma g:
The system $11X^2-6X+1 = 3v^2$ has no positive integer solutions $(X,v)$, and therefore the system $X=2u^2, 11X^2-6X+1 = 3v^2$ has no solutions.
Lemma h:
The only positive integer solution $(X, u, v)$ to the system $X=u^2, 11X^2-6X+1 = 6v^2$ is $(1, 1, 1)$.
I can prove lemmas (a) and (g) fairly easily:
Proof of Lemma a: The equation $X^2-6X+11=v^2$ can be rewritten as $v^2 - (X-3)^2 = 2$. The factors on the left, $v-(X-3)$ and $v+(X-3)$, must have the same parity. However, their product is 2, which is impossible for two integers of the same parity. ∎
Proof of Lemma g: Consider the equation $11X^2-6X+1 = 3v^2$ modulo 4. The LHS can only be congruent to $1$ or $2 \pmod 4$, whereas the RHS can only be congruent to $0$ or $3 \pmod 4$. This contradiction shows there are no integer solutions. ∎
My Question
The remaining six lemmas are unresolved for me. In each lemma, the second equation is a generalized Pell-type equation, whose solutions can be enumerated. However, I am unable to prove that only the specific solutions listed (or none) satisfy the first constraint (e.g., $X=3u^2$ or $X=u^2$).
I tried to imitate the style of proof used in Cohn's theorem on square Fibonacci numbers (related question: Square Fibonacci numbers), but I have not been successful in finding a suitable modulus or other property to restrict the solutions.
Could you please provide any help with the proofs of the remaining lemmas, or suggest any relevant literature or papers that deal with this type of Diophantine problem?
Thank you!
