# 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 $$y$$.

Clio Corvid

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