2 Warmup: general Pythagorean triples

We briefly describe the standard argument [Reference Silverman6, page 13] that uses stereographic projection to find all Pythagorean triples. Although the argument is well known, we include it here because the argument for the Fermat–Mersenne problem follows the same structure, so the simpler problem is worth keeping in mind.

For any Pythagorean triple $(a, b, c)$ , if $d = \gcd (a, b, c)$ , then $(a/d, b/d, c/d)$ is a primitive Pythagorean triple. The Pythagorean triples are precisely the triples of the form $(ka, kb, kc)$ , where k is a positive integer and $(a, b, c)$ is a primitive Pythagorean triple.

In the projective space $\mathbb {P}^{2}$ of points $(X_{0} : X_{1} : X_{2})$ , let C be the curve defined by $X_{0}^{2} = X_{1}^{2} + X_{2}^{2}$ . The intersection of the affine plane $X_{0} \neq 0$ with C is the affine unit circle $1 = X^{2} + Y^{2}$ , where the point $(X, Y)$ in $\mathbb {A}^{2}$ corresponds to the projective point $(X_{0} : X_{1} : X_{2}) = (1 : X : Y)$ in $\mathbb {P}^{2}$ . It is easily checked that the map $(a, b, c) \mapsto (a : b : c)$ from the set of primitive Pythagorean triples to $C(\mathbb {Q})$ is a bijection.

The point $(1 : 0 : 1)$ is on C. In $\mathbb {P}^{2}$ , consider the stereographic projection through $(1 : 0 : 1)$ from C to the projective line $X_{2} = 0$ . For each point P on C, the projection sends P to a point $(t_{0} : t_{1} : 0)$ and we let $f(P)$ be the point $(t_{0} : t_{1})$ in $\mathbb {P}^{1}$ . Calculations show that the map f is given by

$$ \begin{align*} f: C \rightarrow \mathbb{P}^{1}, \quad (X_{0} : X_{1} : X_{2}) \mapsto \left\{\!\!\begin{array}{l} (X_{0} - X_{2} : X_{1}) \\ \quad \text{if }(X_{0} : X_{1} : X_{2}) \neq (1 : 0 : 1) \\ (X_{1} : X_{0} + X_{2}) \\ \quad \text{if } (X_{0} : X_{1} : X_{2}) \neq (1 : 0 : -1) \end{array}\!\!\right\} \end{align*} $$

and that f is an isomorphism over $\mathbb {Q}$ ; its inverse is given by

$$ \begin{align*} f^{-1}: \mathbb{P}^{1} \rightarrow C, \quad (T_{0} : T_{1}) \mapsto (T_{0}^{2} + T_{1}^{2} : 2T_{0}T_{1} : T_{1}^{2} - T_{0}^{2}). \end{align*} $$

(See Section 2 of the supplementary material.) It follows that f restricts to a bijection from $C(\mathbb {Q})$ to $\mathbb {P}^{1}(\mathbb {Q})$ .

Therefore, the map $(a, b, c) \mapsto f((a : b : c))$ from the set of primitive Pythagorean triples to $\mathbb {P}^{1}(\mathbb {Q})$ is a bijection. In the reverse direction, if $t_{0}$ and $t_{1}$ are coprime integers that are not both zero, then it can be checked that the point $(t_{0} : t_{1}) \in \mathbb {P}^{1}(\mathbb {Q})$ corresponds to the primitive Pythagorean triple $(t_{0}^{2} + t_{1}^{2}, 2t_{0}t_{1}, t_{1}^{2} - t_{0}^{2})$ (if $t_{0} + t_{1}$ is odd) or $((t_{0}^{2} + t_{1}^{2})/2, t_{0}t_{1}, (t_{1}^{2} - t_{0}^{2})/2)$ (if $t_{0} + t_{1}$ is even). For a primitive Pythagorean triple $(a, b, c)$ and the corresponding point $(t_{0} : t_{1}) \in \mathbb {P}^{1}(\mathbb {Q})$ , all entries of $(a, b, c)$ are positive if and only if $t_{0} \neq 0$ and $t_{1}/t_{0}> 1$ .

3 The Fermat–Mersenne problem

We now consider the Fermat–Mersenne problem. To be explicit, we define a Fermat–Mersenne tuple to be a 5-tuple $(a, b, c, m, n) \in \mathbb {Z}^{5}$ such that

$$ \begin{align*}a^{2} = b^{2} + c^{2}, \quad a = m^{2}, \quad b + c = n^{2} \quad\text{and}\quad n> 0.\end{align*} $$

To solve the Fermat–Mersenne problem, it is enough to find all Fermat–Mersenne tuples. To obtain the corresponding Pythagorean triples $(a, b, c)$ , we discard m and n. It is easily checked that each Pythagorean triple $(a, b, c)$ obtained in this way comes from exactly two Fermat–Mersenne tuples, which are of the form $(a, b, c, m, n)$ and $(a, b, c, -m, n)$ for some m and n.

For a Fermat–Mersenne tuple $(a, b, c, m, n)$ , if some positive integer p is prime and divides $\gcd (m, n)$ , then $(a/p^{2}, b/p^{2}, c/p^{2}, m/p, n/p)$ is a Fermat–Mersenne tuple. (Note that $p^{4}$ divides $n^{4} - m^{4} = (b + c)^{2} - (b^{2} + c^{2}) = 2bc$ , so $p^{3}$ divides $bc$ , so $p^{2}$ divides b or $p^{2}$ divides c, but $p^{2}$ divides $n^{2} = b + c$ , so $p^{2}$ divides b and $p^{2}$ divides c.) We define a primitive Fermat–Mersenne tuple to be a Fermat–Mersenne tuple $(a, b, c, m, n)$ such that $\gcd (m, n) = 1$ . The Fermat–Mersenne tuples are precisely the tuples of the form $(k^{2}a, k^{2}b, k^{2}c, km, kn)$ , where k is a positive integer and $(a, b, c, m, n)$ is a primitive Fermat–Mersenne tuple.

To find all primitive Fermat–Mersenne tuples, we associate them with the rational points of a certain projective curve and we show that those points correspond to the rational points of a certain elliptic curve.

4 The projective curve

In the space $\mathbb {P}^{3}$ of points given by $(Y_{0} : Y_{1} : Y_{2} : Y_{3})$ , consider the smooth curve C with the two defining equations $Y_{0}Y_{1} = Y_{2}^{2}$ and $2Y_{1}^{2} = Y_{0}^{2} + Y_{3}^{2}$ (see [Reference Silverman6, pages 22–23]). The intersection of the affine 3-space $Y_{0} \neq 0$ with C is given by the equations $X = Y^{2}$ and $2Y^{4} = 1 + Z^{2}$ , where the point $(X, Y, Z)$ in $\mathbb {A}^{3}$ corresponds to the projective point $(Y_{0} : Y_{1} : Y_{2} : Y_{3}) = (1 : X : Y : Z) \in \mathbb {P}^{3}$ . The two points of C that satisfy $Y_{0} = 0$ are the points $(0 : 1 : 0 : \pm \sqrt {2})$ , neither of which is in $C(\mathbb {Q})$ .

Proof. We show the map is well defined. If $(a, b, c, m, n)$ is a primitive Fermat–Mersenne tuple, then the rational numbers $y = m/n$ and $z = (b - c)/n^{2}$ satisfy

$$ \begin{align*} 2y^{4} = \frac{2m^{4}}{n^{4}} = \frac{2a^{2}}{n^{4}} = \frac{2(b^{2} + c^{2})}{n^{4}} = \frac{(b + c)^{2} + (b - c)^{2}}{n^{4}} = 1 + z^{2}, \end{align*} $$

so $(1 : y^{2} : y : z) \in C(\mathbb {Q})$ .

We show the map is injective. Let the point $(Y_{0} : Y_{1} : Y_{2} : Y_{3})$ in $C(\mathbb {Q})$ be the image of some primitive Fermat–Mersenne tuple $(a, b, c, m, n)$ under the map. The integers m and n are uniquely determined by the conditions $n> 0$ , $\gcd (m, n) = 1$ and $Y_{2}/Y_{0} = m/n$ . The integers a, b and c are uniquely determined since $a = m^{2}$ , $b = (n^{2}/2)(1 + (Y_{3}/Y_{0}))$ and $c = (n^{2}/2)(1 - (Y_{3}/Y_{0}))$ .

We show the map is surjective. Let the point $P = (Y_{0} : Y_{1} : Y_{2} : Y_{3})$ be in $C(\mathbb {Q})$ . Now, P must be of the form $(1 : y^{2} : y : z)$ for some rational numbers y and z such that $2y^{4} = 1 + z^{2}$ . Let $(m, n)$ be the unique pair of integers such that $n> 0$ , $\gcd (m, n) = 1$ and $y = m/n$ . Let $d = n^{2}z$ . The number d is in $\mathbb {Q}$ and

$$ \begin{align*}d^{2} = n^{4}z^{2} = n^{4}(2y^{4} - 1) = 2m^{4} - n^{4} \in \mathbb{Z},\end{align*} $$

so d is an integer with the same parity as n. It is easily checked that

$$ \begin{align*}\bigg(m^{2}, \frac{n^{2} + d}{2}, \frac{n^{2} - d}{2}, m, n\bigg)\end{align*} $$

is a primitive Fermat–Mersenne tuple that is sent to P by the map.

For a primitive Fermat–Mersenne tuple $(a, b, c, m, n)$ and the corresponding point $(1 : y^{2} : y : z)$ in $C(\mathbb {Q})$ : n is always positive; $y> 0$ if and only if $m> 0$ , in which case $a = m^{2}> 0$ ; and we have $b> 0$ and $c> 0$ simultaneously if and only if $|z| < 1$ if and only if $|y| < 1$ .

5 The elliptic curve

In the space $\mathbb {P}^{2}$ of points given by $(X_{0} : X_{1} : X_{2})$ , consider the elliptic curve E with the defining equation $X_{0}X_{2}^{2} = X_{1}(X_{1}^{2} + (1/2)X_{0}^{2})$ . The intersection of the affine plane $X_{0} \neq 0$ with E is given by $Y^{2} = X^{3} + (1/2)X$ , where $(X, Y)$ corresponds to the point $(X_{0} : X_{1} : X_{2}) = (1 : X : Y) \in \mathbb {P}^{2}$ . The one point of E that satisfies $X_{0} = 0$ is the point $O = (0 : 0 : 1)$ , which we take as the identity for the group law on E as usual.

The elliptic curve E has complex multiplication by $\mathbb {Z}[i]$ , where the multiply-by-i map

$$ \begin{align*}[i]: E \rightarrow E, \quad (X_{0} : X_{1} : X_{2}) \mapsto (X_{0} : -X_{1} : iX_{2})\end{align*} $$

is an isomorphism over $\mathbb {Q}(i)$ . For each $z \in \mathbb {Z}[i]$ , we write the multiply-by-z map as $[z]: E \rightarrow E: P \mapsto [z]P$ (see [Reference Silverman5, page 107], [Reference Silverman6, pages 69–70] and, more generally, the introductions to the theory of complex multiplication of elliptic curves in [Reference Silverman5, Ch. 2] and [Reference Silverman6, pages 425–429]).

The two points $R = (2 : -1 - i : 1 - i)$ and $(1 : 0 : 0)$ are on E, and they satisfy $[2]R = (8 : 2 : 3)$ and $[2](1 : 0 : 0) = O$ . Each of the two maps

$$ \begin{align*}\begin{array}{c} \mathbb{Z} \times (\mathbb{Z}/2\mathbb{Z}) \rightarrow E(\mathbb{Q}),\quad (t, u + 2\mathbb{Z}) \mapsto [t](8 : 2 : 3) + [u](1 : 0 : 0), \\\mathbb{Z} \times \mathbb{Z} \times (\mathbb{Z}/2\mathbb{Z}) \rightarrow E(\mathbb{Q}(i)),\quad (s, t, u + 2\mathbb{Z}) \mapsto [s + ti]R + [u](1 : 0 : 0) \end{array}\end{align*} $$

is a group isomorphism. (A proof may be given using the techniques of [Reference Silverman6, Example 4.5 in Section III.4, Sections VII.1–3 and Proposition 4.9 in Section X.4]. This may also be checked using SageMath and a database like LMFDB; the elliptic curve E is isomorphic to the curve with Cremona label 256b2 [7]. See Section 3 of the supplementary material.)

6 Correspondences between the two curves

The point $(1 : 0 : 0 : i)$ is on C. Starting with a point $P = (y_{0} : y_{1} : y_{2} : y_{3})$ on C, apply stereographic projection in $\mathbb {P}^{3}$ through $(1 : 0 : 0 : i)$ from C to the projective plane $Y_{3} = 0$ . We obtain a point $(x_{0} : x_{1} : x_{2} : 0)$ ; let this correspond to the point $f(P) = (x_{0} : x_{1} : x_{2})$ in $\mathbb {P}^{2}$ .

Proof. Calculations show that f sends each point in C to a point in E, that f is given by

$$ \begin{align*}f: C \rightarrow E, \quad (Y_{0} : Y_{1} : Y_{2} : Y_{3}) \mapsto \left\{\!\!\begin{array}{l} (Y_{0} + iY_{3} : Y_{1} : Y_{2})\\ \quad \text{if } (Y_{0} : Y_{1} : Y_{2} : Y_{3}) \neq (1 : 0 : 0 : i)\\ (2Y_{1}Y_{2} : Y_{2}(Y_{0} - iY_{3}) : Y_{0}(Y_{0} - iY_{3}))\\ \quad \text{if } (Y_{0} : Y_{1} : Y_{2} : Y_{3}) \text{ is neither}\\ \quad (1 : 0 : 0 : -i) \text{ nor } (0 : 1 : 0 : \pm\sqrt{2}) \end{array}\!\!\right\}\end{align*} $$

and that the map

$$ \begin{align*}f^{-1}: E \rightarrow C, \quad (X_{0} : X_{1} : X_{2}) \mapsto \left\{\!\!\begin{array}{l} (2X_{1}^{2} + X_{0}^{2}: 2X_{0}X_{1} : 2X_{0}X_{2}: i(2X_{1}^{2} - X_{0}^{2}))\\ \quad \text{if } (X_{0} : X_{1} : X_{2}) \neq (0 : 0 : 1)\\ (X_{2}^{2}: X_{1}^{2} : X_{1}X_{2} : i(X_{2}^{2} - X_{0}X_{1}))\\ \quad \text{if } (X_{0} : X_{1} : X_{2}) \neq (1 : 0 : 0) \end{array}\!\!\right\}\end{align*} $$

is the inverse of f. (See Sections 4.1–4.3 of the supplementary material.) The maps f and $f^{-1}$ are defined over $\mathbb {Q}(i)$ , and the result follows. (The map f can be obtained from the map $\phi ^{-1}$ in [Reference Silverman6, pages 327–329] with $d = -1$ , $a = 0$ and $b = 1/2$ by rearranging and renaming co-ordinates, post-composing with $[i]$ and taking the projective closure where necessary.)

Table 1 lists some points on E and on C that correspond under f. It is easily checked that the maps

$$ \begin{align*}\begin{array}{l} q: C \rightarrow C, \quad (Y_{0} : Y_{1} : Y_{2} : Y_{3}) \mapsto (Y_{0} : Y_{1} : -Y_{2} : -Y_{3}), \\ r: C \rightarrow C, \quad (Y_{0} : Y_{1} : Y_{2} : Y_{3}) \mapsto (Y_{0} : -Y_{1} : iY_{2} : Y_{3}) \end{array}\end{align*} $$

are isomorphisms over $\mathbb {Q}(i)$ , and that every point P on C satisfies

$$ \begin{align*}f(P) + (1 : 0 : 0) = f(q(P)) \quad\text{and}\quad [i](f(P)) = f(r(P)).\end{align*} $$

(See Section 4.4 of the supplementary material.)

Table 1

Points on E and points on C that correspond under f. The point R is the point $(2 : -1 - i : 1 - i)$ in $E(\mathbb {Q}(i))$ . Each point that is defined over $\mathbb {Q}$ is marked with an asterisk.

Table 1 Long description

The table consists of four columns and thirteen data rows.

Column headers:

1. s

2. t

3. [s + ti]R in E

4. f super -1 ([s + ti]R) in C

Data rows:

- Row 1: s = 0, t = -3. Column 3: (4394 : 13 - 3107i : 1 - 239i). Column 4: * (1 : 169 : 13 : 239).

- Row 2: s = 0, t = -2. Column 3: (8 : -2 : -3i). Column 4: (9 : -4 : -6i : -7i).

- Row 3: s = 0, t = -1. Column 3: (2 : 1 + i : -1 - i). Column 4: * (1 : 1 : -1 : -1).

- Row 4: s = -3, t = 0. Column 3: (4394 : -13 + 3107i : -239 - i). Column 4: (1 : -169 : -13i : 239).

- Row 5: s = -2, t = 0. Column 3: * (8 : 2 : -3). Column 4: (9 : 4 : -6 : -7i).

- Row 6: s = -1, t = 0. Column 3: (2 : -1 - i : -1 + i). Column 4: (1 : -1 : i : -1).

- Row 7: s = 0, t = 0. Column 3: * (0 : 0 : 1). Column 4: (1 : 0 : 0 : i).

- Row 8: s = 1, t = 0. Column 3: (2 : -1 - i : 1 - i). Column 4: (1 : -1 : -i : -1).

- Row 9: s = 2, t = 0. Column 3: * (8 : 2 : 3). Column 4: (9 : 4 : 6 : -7i).

- Row 10: s = 3, t = 0. Column 3: (4394 : -13 + 3107i : 239 + i). Column 4: (1 : -169 : 13i : 239).

- Row 11: s = 0, t = 1. Column 3: (2 : 1 + i : 1 + i). Column 4: * (1 : 1 : 1 : -1).

- Row 12: s = 0, t = 2. Column 3: (8 : -2 : 3i). Column 4: (9 : -4 : 6i : -7i).

- Row 13: s = 0, t = 3. Column 3: (4394 : 13 - 3107i : -1 + 239i). Column 4: * (1 : 169 : -13 : 239).

Note: Asterisks indicate points defined over the field Q.

We would like to find an isomorphism from C to E over $\mathbb {Q}$ instead of over $\mathbb {Q}(i)$ so that points in $C(\mathbb {Q})$ correspond to points in $E(\mathbb {Q})$ . Table 1 suggests that we consider the map $g: C \rightarrow E, P \mapsto [-i](f(P)) - R$ (which sends the rational points $(1 : 1 : 1 : -1)$ and $(1 : 169 : -13 : 239)$ on C to respectively the rational points O and $(8 : 2 : 3)$ on E).

Proof. By Proposition 6.1, the map g is an isomorphism over $\mathbb {Q}(i)$ and its inverse is $g^{-1}: E \rightarrow C,\ P \mapsto f^{-1}([i](P + R))$ . Calculations yield the formulae for g and $g^{-1}$ :

$$ \begin{align*}\begin{array}{l} g: C \rightarrow E,\quad (Y_{0} : Y_{1} : Y_{2} : Y_{3}) \mapsto (X_{0} : X_{1} : X_{2})\\ = \left\{\!\!\begin{array}{l} ((-Y_{0} + 2Y_{1} + Y_{3})^{2} : (Y_{0} + Y_{1} + 2Y_{2})(-Y_{0} + 2Y_{1} + Y_{3}) :\\ Y_{0}^{2} + Y_{0}Y_{2} + Y_{0}Y_{3} + 2Y_{1}^{2} + 2Y_{1}Y_{2} + 2Y_{1}Y_{3} + 3Y_{2}Y_{3})\\ \qquad \text{if } (Y_{0} : Y_{1} : Y_{2} : Y_{3}) \text{ is neither } (1 : 1 : 1 : -1) \text{ nor } (1 : 1 : -1 : -1) \\ (2(Y_{2} + Y_{3})(Y_{0} + Y_{1} - 2Y_{2}) : -(Y_{2} + Y_{3})(Y_{0} - 2Y_{1} + Y_{3}) : \\ -(Y_{0} + 2Y_{1})(Y_{0} - 2Y_{1} + Y_{3})) \quad \text{if } (Y_{0} : Y_{1} : Y_{2} : Y_{3}) \text{ is not among} \\ \qquad (1 : 1 : \pm 1 : 1), (2 : -1 : i\sqrt{2} : -i\sqrt{2}) \text{ and } (2 : -1 : -i\sqrt{2} : i\sqrt{2}) \end{array}\!\!\right\}, \end{array}\end{align*} $$

$$ \begin{align*}\begin{array}{l} g^{-1}: E \rightarrow C,\quad (X_{0} : X_{1} : X_{2}) \mapsto (Y_{0} : Y_{1} : Y_{2} : Y_{3})\\ = \left\{\!\!\begin{array}{l} \begin{array}{l} ((X_{2}^{2} + X_{1}X_{2} - X_{0}X_{1})^{2} : (X_{1}^{2} + X_{2}^{2})^{2} : (X_{1}^{2} + X_{2}^{2})(X_{2}^{2} + X_{1}X_{2} - X_{0}X_{1}) :\\ (X_{1}^{2} - X_{2}^{2})(X_{2}^{2} + X_{1}X_{2} - X_{0}X_{1}) + 3X_{1}X_{2}(X_{1}^{2} + X_{2}^{2})) \end{array}\\ \qquad \text{if } (X_{0} : X_{1} : X_{2}) \text{ is not among } (1 : 0 : 0), R \text{ and } (1 : 0 : 0) + R\\ \begin{array}{l} ((X_{0}^{2} - 4X_{0}X_{1} + 2X_{1}^{2})^{2} : (-X_{0}^{2} + 2X_{1}^{2} - 2X_{0}X_{2})^{2} :\\ (X_{0}^{2} - 4X_{0}X_{1} + 2X_{1}^{2})(-X_{0}^{2} + 2X_{1}^{2} - 2X_{0}X_{2}) :\\ (X_{0}^{2} - 2X_{1}^{2} + 2X_{0}X_{2})(X_{0}^{2} + 4X_{0}X_{1} + 2X_{1}^{2}) + 6X_{0}X_{2}(X_{0}^{2} - 4X_{0}X_{1} + 2X_{1}^{2})) \end{array}\\ \qquad \text{if } (X_{0} : X_{1} : X_{2}) \text{ is not among } O, (2 : i\sqrt{2} : 0) + R \text{ and }\\ \qquad (2 : -i\sqrt{2} : 0) + R\\ \end{array}\!\!\right\}. \end{array}\end{align*} $$

(See Sections 4.5 and 4.6–4.7 of the supplementary material, respectively.)

The maps g and $g^{-1}$ are defined over $\mathbb {Q}$ , and the result follows.