*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 \(y\).

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