Gerono: Q. 1177

This is my translation of Gerono’s 1877 proof listing all the possible solutions (x, y) for the equation $y^2 = x^3 + x^2 + x + 1$.

“Solutions to questions posed in The New Annals: Question 1177.” MM. Gerono, Nouvelles annales de mathématiques. Journal des candidats aux écoles polytechnique et normale, second series, volume 16 (1877), 230-234 (See second series, volume 14, p. 288)

Find all positive integer solutions to the equation $$y^2 = x^3 + x^2 + x + 1$$ (Brocard)

Author’s note: I have addressed two solutions based on different calculations; they lead to the values 1 and 7 for the unknown $x$. However, in neither of these cases has it been rigorously demonstrated that $x$ does not allow for other positive integer values, which remains to be seen.

In order to find all integer values of $x$, both positive and negative, I suppose in order that $x < 0$, $x = 0$, and $x > 0$.

When $x$ is negative, we can replace $x$ with $-z$, which produces $$y^2 = -z^3 + z^2 – z + 1 \\ (1-z)(1+z^2)=y^2 \\ z = 1, y = 0 \\ x=-1, y = 0$$

If $x = 0$, then $y = \pm 1$.

The hypotheses of $x < 0$ and $x = 0$ not having any other solutions, I will now consider the inequality $x > 0$, as follows.

The original equation can be written $$(x + 1)(x^2 + 1) = y^2$$

The integers represented by $(x + 1)$ and $(x^2 + 1)$ must have a common factor other than one because, if they were co-prime, then $x^2 + 1$ would be a square number, which is obviously impossible.

The common factor is two, which follows from the identity $x^2 + 1 = (x + 1)(x – 1) + 2$.

It follows that both of the numbers $(x + 1)$ and $(x^2 + 1)$ must be double a square number, which leads to the equations of the form: $$x + 1 = 2m^2 \\ x^2 + 1 = 2n^2$$

Solving each equation for $x$ and then eliminating it leads to: $$m^4 + (m^2 – 1)^2 = n^2$$ which shows that $n$ is an odd number, first with $m$ and then with $m^2 – 1$.

This posited, I distinguish two cases, depending on whether $m$ is odd or even.

First: $m$ is odd. The most recent equation can be rewritten as $$[n + (m^2 – 1)][n – (m^2 – 1)] = m^4$$

The odd numbers $n + (m^2 – 1)$ and $n – (m^2 – 1)$ are co-prime, since their sum ($2n$) and their difference ($2(m^2 – 1)$) do not have a common factor other than two. Thus each of these numbers is an integer to the fourth power. Let: $$n + (m^2 – 1) = p^4$$ and $$n – (m^2 – 1) = q^4$$ where $$p^4q^4 = m^4; p^2q^2 = m^2$$

These two equations give us $$p^4 – q^4 = 2(m^2 – 1)$$ where, due to $p^2q^2 = m^2$, $$p^4 – q^4 = 2(p^2q^2 – 1) \\ q^4 + 2p^2q^2 – (p^4 + 2) = 0 \\ q^2 = – p^2 + \sqrt{2(p^4 + 1)}$$

We can see from this last equation that $p^4 + 1$ is twice a square, which requires that $p = 1$ (Legendre, Théorie des nombres, volume two, pp. 4-5, 1830 edition: “The equation $x^4 + y^4 = 2p^2$ is impossible unless $x = y$.”). Thus, $$q^2 = -1 + \sqrt{4} = 1, m^2 = 1, x = 1, \text{ and } y = \pm 2$$

Hence, if $m$ is odd, the solutions of the equation $y^2 = x^3 + x^2 + x + 1$ are, in their entirety, $x = 1, y = 2$ and $x = 1, y = -2$.

Second: $m$ is even. In the equation $$[n + (m^2 – 1)][n – m^2 – 1)] = m^4$$ the factors $n + (m^2 – 1)$ and $n – (m^2 – 1)$ currently represent even numbers. In letting: $$n + (m^2 – 1) = 2a \\ n – (m^2 – 1) = 2b \\ m = 2r$$ we have $$n = a + b \\ m^2 – 1 = a – b \\ 4ab = m^4 = 16r^4 \\ ab = 4r^4$$

$n$ and $m^2 – 1$ are co-prime, and it is the same with $a$ and $b$. Consequently, by virtue of the equality $ab = 4r^4$, one of the two numbers $(a, b)$ is four times an integer to the fourth power and the other is an integer to the fourth power; that is, we have: $$a = 4p^4, b = q^4 \text{ or } a = p^4, b = 4q^4$$

But the equalities $a = p^4, b = 4q^4$ are not acceptable, because they result in $m^2 – 1 = p^4 – 4q^4$, an absurd equality in which the first member is one less than a multiple of four and the other is one more than a multiple of four. We must therefore take $$a = 4p^4, b = q^4$$

It follows that $$m^2 – 1 = 4p^4 – q^4$$ and that $$4p^4q^4 = 4r^4, 4p^2q^2 = 4r^2 = m^2$$ where $$4p^2q^2 – 1 = 4p^4 – q^4 \\ q^4 + 4p^2q^2 – (4p^4 + 1) = 0 \\ q^2 = -2p^2 + \sqrt{8p^4 + 1}$$

$8p^4 + 1$ being necessarily equal to the square of an odd number, we have $$8p^4 + 1 = (2k + 1)^2$$ and, as a result, $$\frac{k(k+1)}{2} = p^4$$ an equation which yields $$p^2 = 1 \text{ or } p = 0$$ (Legendre, Théorie des nombres, volume two, p. 7, 1830 edition: “No triangular number $\frac{x(x+1)}{2}$ except 1 is equal to an integer to the fourth power.”)

For $p^2 = 1, q^2 = -2 + \sqrt{9} = 1$, $$m^2 = 4, x = 7, y = \pm 20$$ and for $p^2 = 0, q^2 = 1,$ $$m^2 = 0, x = -1, y = 0$$

Therefore, when $m$ is even, the complete list of solutions of the equation $y^2 = x^3 + x^2 + x + 1$ are $$x = 7, y = \pm 20 \text{ and  } x = -1, y = 0$$ This last solution was already mentioned.

Based on all of the preceding, we conclude that the indeterminate equation $y^2 = x^3 + x^2 + x + 1$ has the solutions $$x = -1, y = 0 \\ x = 0, y = \pm 1 \\ x = 1, y = \pm 2 \\ x = 7, y = \pm 20$$ and that it is not possible to find any other values for the unknowns $x$ and $y$.