Proof. Let $X, Y$ satisfying $X^4+Y^4=2^m$ lie in a quadratic number field K. Suppose that $T \in \mathbb {Q}$ and that $u = X/Y \in \mathbb {Q}$ . Then, using (1.2), we deduce that $u^2 {\kern-1.5pt}={\kern-1.5pt} (T^2 {\kern-1.5pt}+{\kern-1.5pt}2T-1)/ {(T^2 {\kern-1.5pt}-{\kern-1.5pt} 2T {\kern-1.5pt}-{\kern-1.5pt} 1)}$ . Setting $v {\kern-1.5pt}={\kern-1.5pt} u(T^2 {\kern-1.5pt}-{\kern-1.5pt} 2T {\kern-1.5pt}-{\kern-1.5pt} 1) {\kern-1.5pt}\in{\kern-1.5pt} \mathbb {Q}$ yields ${v^2 {\kern-1.5pt}={\kern-1.5pt} T^4 {\kern-1.5pt}-{\kern-1.5pt} 6T^2 {\kern-1.5pt}+{\kern-1.5pt} 1}$ , which is a curve with rational point $(T, v) \in \mathbb {Q} \times \mathbb {Q}$ . Following the proof of [Reference Mordell9, Theorem 2 on page 77], we make a change of variables. First, we set ${v {\kern-1.5pt}={\kern-1.5pt} T^2 {\kern-1.5pt}-{\kern-1.5pt} 1 {\kern-1.5pt}-{\kern-1.5pt} 2x}$ . Note that $x \neq 1$ , because if $x = 1$ , then we have $v = T^2 - 3$ , which would contradict $v^2 = T^4 - 6T^2 + 1$ . Next, we set $T = y / (1-x)$ . In this way, the point $(T, u) \in \mathbb {Q} \times \mathbb {Q}$ yields a rational point $(x, y)$ with $x \neq 1$ on the elliptic curve $y^2 = x^3 - x$ , which has LMFDB label 32.a3 and Mordell–Weil group $\mathbb {Z}/2\mathbb {Z} \oplus \mathbb {Z}/2\mathbb {Z}$ . This elliptic curve has three rational points $(0, 0), (1, 0), (-1, 0)$ , but only $(x, y) = (0, 0), (-1, 0)$ satisfy $x \neq 1$ . Since $y = 0$ in these two cases, we find that $T = 0$ . Since $T = 0$ , we have $u^2 = 1$ , which means that $X^2 = Y^2$ and thus $X^4 = Y^4$ .

Assume now that $X^4 = Y^4$ . Then, $X^4 + Y^4 = 2$ reduces to $X^4 = 1$ and thus ${K = \mathbb {Q}(\sqrt {-1})}$ . Similarly, $X^4 + Y^4 = 8$ reduces to $X^4 = 4$ and thus $\mathbb {Q}(\sqrt {\pm 2})$ .

Proof. We first begin with $m = 1$ and the curve $X^4 + Y^4 = 2$ . Without loss of generality, $(X, Y) = (a, b_1 \sqrt {n})$ and, thus, $a^4 + b_1^4 n^2 = 2$ gives $j^2 + k^4 = 2$ . Notice that if $k = 1$ , then $j = \pm 1$ , which contradicts $X^4 \neq Y^4$ . Thus, $k \neq 1$ , and using [Reference Mordell9, proof of Theorem 2 on page 77] and the solution $(j, k) = (1, 1)$ to $j^2 + k^4 = 2$ , we make the change of variables $j = v(w+1)^{-2}$ , $k= 1 + (w+1)^{-1}$ , which yields $v^2 = w^4 - 12w^2 - 24w - 14$ . Since $k \neq 1$ , the rational point $(j, k)$ produces a rational point $(w, v) \in \mathbb {Q} \times \mathbb {Q}$ . We then set $v = -w^2 +2h + 2$ to determine that

$$ \begin{align*}w^2 - \frac{6}{h-2}w = \frac{h^2 + 2h + {9}/{2}}{h-2}.\end{align*} $$

Completing the square for the quadratic in w determines that $((h-2)w - 3)^2 = h^3 + h/2$ . Replacing $(h-2)w - 3$ with $y/8$ and letting $x = 4h$ yields a rational point on the elliptic curve $y^2 = x^3 + 8x$ with LMFDB label 256.b2, which has Mordell–Weil group $\mathbb {Z} \oplus \mathbb {Z}/2\mathbb {Z}$ with the point $(1, 3)$ of infinite order.

Working backwards from any rational point $(a, b)\neq (8,-24)$ on the elliptic curve $y^2 = x^3 + 8x$ to our curve $X^4 + Y^4 = 2$ yields the solution

$$ \begin{align*}\bigg(\frac{4a+b-8}{2a+b+8} \bigg)^4 + \bigg(\frac{\sqrt{2a^3-24a^2-b^2 - 48b - 64}}{2a+b+8} \bigg)^4 = 2\end{align*} $$

in the field $K=\mathbb {Q} (\sqrt {2a^3 - 24a^2 - b^2 - 48b-64})$ . This field is quadratic, namely ${K\neq \mathbb {Q}}$ , for every rational point $(a,b)$ on $y^2 = x^3 + 8x$ with the exception of $(a,b)=(1,-3)$ . Note that for $(a,b)=(1,-3)$ , we have $2a^3 - 24a^2 - b^2 -48b-64=49$ and thus ${K = \mathbb {Q}}$ . To prove this field is otherwise quadratic, we use the fact that, as Lebesgue proved, $(X, Y)=(\pm 1,\pm 1)$ are the only rational solutions to $X^4 + Y^4 = 2$ . Set

$$ \begin{align*} A(x,y) = \frac{4x+ y - 8}{2x + y+ 8}, \quad B(x,y) = \frac{\sqrt{2x^3 - 24x^2 - y^2 - 48y - 64}}{2x + y + 8}. \end{align*} $$

Suppose that $B(a,b)$ is rational. This makes $(A(a, b),B(a,b))$ a rational point on $X^4 + Y^4 = 2$ . Thus, $A(a,b)=\pm 1$ which, using $y^2=x^3+8x$ , means that $(a,b)=(0,0), (1,-3), (8,24)$ . Recall that the point $(8,-24)$ was excluded. A quick calculation shows that $B(a,b)\neq \pm 1$ for $(a,b)=(0,0)$ and $(a, b) = (8,24)$ . We record that $(A(1,-3),B(1,-3))=(-1,1)$ .

We now use Falting’s theorem and the fact that the Fermat quartic ${X^4+Y^4=2}$ is a nonsingular curve of genus 3 to prove there are infinitely many different quadratic fields $\mathbb {Q}(\sqrt {2a^3 - 24a^2 -b^2 - 48b-64}) \not = \mathbb {Q}$ , where $(a, b)$ is a rational point on ${y^2 = x^3 + 8x}$ . Since each quadratic field K contains only finitely many solutions to ${X^4+Y^4=2}$ , this will follow once we establish the fact that there are infinitely many different complex ordered pairs $(A(a,b),B(a, b))\in \mathbb {C}\times \mathbb {C}$ .

Since $A(a,b)\in \mathbb {Q}$ and $B(a_1,b_1)\not \in \mathbb {Q}$ for all rational points $(a_1, b_1)$ on ${y^2 = x^3 + 8x}$ except $(a_1, b_1)=(1,-3)$ , there is no possibility of $A(a,b)=B(a_1,b_1)$ unless ${B(a_1,b_1)=1}$ . If $B(a_1,b_1)=1$ , then $A(a_1,b_1)=\pm 1$ since $A(a_1,b_1)$ is rational. Recall that we have already described the finitely many points $(a_1, b_1)$ on $y^2=x^3+8x$ such that $A(a_1, b_1)=\pm 1$ . Thus, we focus on proving there are infinitely many different ordered pairs $(A(a, b),B(a,b))\in \mathbb {C}\times \mathbb {C}\setminus \{(\pm 1,1)\}$ .

Our only concern is the possibility that there might be infinitely many rational points $(a_i, b_i)$ on $y^2 = x^3 + 8x$ such that $(A(a_i,b_i),B(a_i,b_i))=(A(a_j,b_j),B(a_j,b_j))\not \in \{(\pm 1,1)\}$ . If this were the case, letting $k=A(a_i,b_i)\in \mathbb {Q}$ be the common value of the $A(a_i,b_i)$ , we would have infinitely many intersections of $y^2 = x^3 + 8x$ with the level curve

$$ \begin{align*}k=\frac{4x+y-8}{2x+y+8}.\end{align*} $$

Recall that since $(A(a,b),B(a,b))\not \in \{(\pm 1,1)\}$ , we have $k\neq \pm 1$ . Since $k\neq 1$ , we may solve for y in the equation for the level curve and then use this to substitute for y in $y^2 = x^3 + 8x$ . The result is a cubic equation in x with no more than three rational roots. As a result, the infinitely many rational points on $y^2 = x^3 + 8x$ produce infinitely many different quadratic extensions K.

We now consider the case $m = 3$ and the curve $X^4 + Y^4 = 8$ . Again without loss of generality, we have $(X, Y) = (a, b_1\sqrt {n})$ with $a,b_1\in \mathbb {Q}$ . Then, $a^4 + b_1^4 n^2 = 8$ and thus a nontrivial integer solution for $8x^4 - y^4 = z^2$ . However, this curve is known to have no nontrivial integer solutions by Lebesgue [Reference Lebesgue6] using the method of infinite descent, and thus there are no nontrivial quadratic fields that come out of this case.

Proof. We begin by supposing that $T\not \in \mathbb {Q}$ and $T^2\in \mathbb {Q}$ . Thus, we have $T = r\sqrt {n}\not \in \mathbb {Q}$ for some nonzero $r\in \mathbb {Q}$ . Using (1.2), this means that $\sigma (X^2)=Y^2$ , where $\sigma :K\rightarrow K$ denotes conjugation. Since $\sigma $ is a multiplicative homomorphism, $\sigma (X)^2=Y^2$ , which means that $Y=\pm \sigma (X)$ and $(X, Y)$ is a quadratic conjugate pair.

Now, we derive contradictions in the more complicated case in which $T^2\not \in \mathbb {Q}$ . Following the argument Mordell used in [Reference Mordell8], set $Z = (1+T^2)XY \in K$ . Using (1.2), we find that $Z^2 = 2^{m-1}(T^4 - 6T^2 + 1)$ . Set $V = Z/(2^{(m-1)/2}) - T^2 + 3 \in K$ . Since $Z = 2^{(m-1)/2} (V + T^2 -3)$ , we have $V^2 + 2VT^2 - 6V + 8 = 0$ . If $V \in \mathbb {Q}$ , then because $V^2 + 2VT^2 - 6V + 8 = 0$ , we have $T^2 \in \mathbb {Q}$ , which is a contradiction. Thus, $V \not \in \mathbb {Q}$ , and there exist $c, d\in \mathbb {Q}$ with $d \neq 0$ such that $V = c+dT$ . Define the linear polynomial ${V(z) = c + dz \in \mathbb {Q}{[z]}}$ and observe that the polynomial

$$ \begin{align*}p(z) = V(z)^2 + 2V(z)z^2 - 6V(z) + 8 \in \mathbb{Q}{[z]}\end{align*} $$

has T as a root. Since $d \neq 0$ , this polynomial is cubic, which means that ${p(z) = F(z) G(z)}$ , where $F(z) \in \mathbb {Q}{[z]}$ is the monic irreducible quadratic polynomial for T and ${G(z) \in \mathbb {Q}[z]}$ is linear. At this point,

(3.1) $$ \begin{align} V(z)^2 + 2V(z)z^2 - 6V(z) + 8 = F(z)G(z). \end{align} $$

Since $G(z)$ is linear, it has a rational root e, which means that $(x, y) = (e, V(e))$ is a rational point on the curve $y^2 + 2x^2y - 6y + 8 = 0$ . This elliptic curve has minimal Weierstrass equation $y^2 = x^3 - x$ with LMFDB label 32.a3 and Mordell–Weil group $\mathbb {Z}/2\mathbb {Z}\oplus \mathbb {Z}/2\mathbb {Z}$ . The four rational projective points on $y^2z + 2x^2y - 6yz^2 + 8z^3 = 0$ are $(x : y : z)=(0: 1:0), (0:2:1), (0:4:1)$ and $(1:0:0)$ . Thus, we determine that either $(e, V(e)) = (0, 2)$ or $(e, V(e)) =(0, 4)$ . In both of these cases, the root of the linear polynomial $Q(z) $ is $0$ ; therefore, $G(z) = hz$ for some $h \in \mathbb {Q}$ . Thus, (3.1) becomes $(c+dz)^2 + 2(c +dz)z^2 - 6(c+dz) + 8 = F(z)hz.$ Comparing the leading terms, we see that $h=2d$ . Comparing the constant terms, we see that $c^2 - 6c + 8 = 0$ . Thus, $c = 2$ or $4$ , and solving for $F(z)$ , we find that

$$ \begin{align*}F(z) = z^2 + z(2c+d^2)/2d + (c-3),\end{align*} $$

which means that

(3.2) $$ \begin{align} T=\frac{-d^2-2c\pm\sqrt{d^4+(48-12c)d^2+4c^2}}{4d}.\end{align} $$

There are two cases $c = 2$ and $c = 4$ to consider. For $c=2$ , substituting T into (1.2) yields

$$ \begin{align*}X^2 = \pm\frac{2^{{(m-1)}/{2}}(d-2)^2}{\sqrt{d^4 + 24d^2 + 16}}, \quad Y^2 = \pm\frac{2^{{(m-1)}/{2}}(d+2)^2}{\sqrt{d^4 + 24d^2 + 16}},\end{align*} $$

where the $\pm $ signs agree with the $\pm $ sign in (3.2). Note that $X^4,Y^4\in \mathbb {Q}$ . If either of $X^2$ or $Y^2\in \mathbb {Q}$ , then both are rational and $T\in \mathbb {Q}$ , which contradicts $T^2\not \in \mathbb {Q}$ . Thus, we find that $X^4,Y^4\in \mathbb {Q}$ , while neither $X^2\in \mathbb {Q}$ nor $Y^2\in \mathbb {Q}$ . Letting $X=r+s\sqrt {n}$ with $r,s\in \mathbb {Q}$ , and expanding $X^2$ and $X^4$ , yields $rs\neq 0$ and $rs(r^2+s^2n)=0$ . So $n=-r^2/s^2$ . Since n is a squarefree integer, $n=-1$ and $s=r$ . Thus, $X=r(1+\sqrt {-1})$ and $X^4=-4r^4$ . Similarly, $Y^4=-4r_1^4$ for some $r_1\in \mathbb {Q}$ . However, this means that $X^4+Y^4 < 0$ , which contradicts $X^4+Y^4=2^m$ .

For $c=4$ , we substitute the values for T into (1.2) separately. In the first place, for $T=(-d^2-8+\sqrt {d^2+64})/4d$ , we find

$$ \begin{align*}X^2 = 2^{{(m-1)}/{2}}\cdot\frac{-4d - \sqrt{d^4 + 64} }{d^2 + 8}, \quad Y^2 = 2^{{(m-1)}/{2}}\cdot\frac{4d - \sqrt{d^4 + 64} }{d^2 + 8}.\end{align*} $$

Second, for $T=(-d^2-8-\sqrt {d^2+64})/4d$ , we find

$$ \begin{align*}X^2 = 2^{{(m-1)}/{2}}\cdot\frac{-4d + \sqrt{d^4 + 64} }{d^2 + 8}, \quad Y^2 = 2^{{(m-1)}/{2}}\cdot\frac{4d + \sqrt{d^4 + 64} }{d^2 + 8}.\end{align*} $$

Since Lebesgue [Reference Lebesgue6] proved that the Fermat quartic $x^4+2^my^4=z^2$ has only nonzero integer solutions for $m \equiv 2 \pmod {4}$ and $d\in \mathbb {Q}$ is nonzero, $\sqrt {d^4 + 64}\not \in \mathbb {Q}$ . Thus, ${K=\mathbb {Q}(\sqrt {d^4+64})}$ is a real-valued quadratic field. Consider the square of any element of K, namely $(a+b\sqrt {d^4+64})^2=a^2+b^2(d^4+64)+2ab\sqrt {d^4+64}$ . Notice that the rational summand is positive. This leads to a contradiction in the expressions for $X^2$ and $Y^2$ , since $-4d$ and $4d$ have opposite signs.

Proof. We begin with $m = 1$ and suppose there exists a solution $(X,Y)\in K\times K$ to $X^4+Y^4=2$ with $X=a+b\sqrt {n}$ , $Y=\pm (a-b\sqrt {n})$ for some $a,b\in \mathbb {Q}$ . Expanding $(a+ b\sqrt {n}) ^4 + (a - b\sqrt {n})^4 = 2$ produces $a^4 + 6a^2b^2n + b^4n^2 = 1$ . The change of variable $b^2n = c- 3a^2$ yields $c^2 - 8a^4 = 1$ . Then, the further change of variables

$$ \begin{align*}(a, c) = \bigg( \frac{x}{y}, \frac{y^2 - 2x^3}{y^2} \bigg) \implies b^2n = \frac{y^2 - 2x^3 - 3x^2}{y^2} = \frac{-x^3 - 3x^2 - 2x}{y^2}\end{align*} $$

yields a rational point on the the elliptic curve $y^2 = x^3 - 2x$ with LMFDB label 256.b1, Mordell–Weil group $\mathbb {Z} \oplus \mathbb {Z}/2\mathbb {Z}$ with point $(2, 2)$ of infinite order.

Working backwards from any rational point $(a,b)\neq (0,0)$ on the elliptic curve ${y^2 = x^3 - 2x}$ to $X^4 + Y^4 = 2$ yields the solution

$$ \begin{align*}\bigg(\frac{a + \sqrt{-a^3 -3a^2 -2a}}{b} \bigg)^4 + \bigg(\frac{a - \sqrt{-a^3 - 3a^2 - 2a}}{b} \bigg)^4 = 2\end{align*} $$

in the field $K=\mathbb {Q} (\sqrt {-a^3 - 3a^2 - 2a})$ . Since, as Lebesgue proved, there are no nontrivial rational points on $X^4 + Y^4 = 2$ , it follows that $\sqrt {-a^3 - 3a^2 - 2a}\not \in \mathbb {Q}$ and K is a quadratic extension of $\mathbb {Q}$ .

We now use Falting’s theorem and the fact that $X^4+Y^4=2$ is a nonsingular curve of genus 3 to prove there are infinitely many different quadratic fields $\mathbb {Q} (\sqrt {-a^3 - 3a^2 - 2a})$ with $(a, b)$ a rational point on the elliptic curve $y^2 = x^3 - 2x$ . Since each quadratic field K contains only finitely many solutions to $X^4+Y^4=2$ , this result will follow once we establish the fact that there are infinitely many different values of

$$ \begin{align*}A(x,y)=\frac{\sqrt{-x^3 - 3x^2 - 2x}}{y}\in\mathbb{C}.\end{align*} $$

Our concern is that there might be simultaneous values $k=A(x,y)$ that occur for infinitely many different points $(a,b)$ on $y^2 = x^3 - 2x$ . So we ask whether there can be any complex values $k\in \mathbb {C}$ such that the level curve $y^2k^2=-x^3 - 3x^2 - 2x$ intersects $y^2=x^3-2x$ at infinitely many real points $(x_1,y_1)$ . Since each intersection point yields a real solution to the polynomial $k^2(x^3-2x)=-x^3 - 3x^2 - 2x$ , which is a polynomial with no more than three distinct roots, this does not occur.

Now, we let $m = 3$ and suppose there exists a solution $(X,Y)\in K\times K$ to the curve $X^4+Y^4=8$ with $X=a+b\sqrt {n}$ , $Y=\pm (a-b\sqrt {n})$ for some $a,b\in \mathbb {Q}$ . Expanding $(a+ b\sqrt {n}) ^4 + (a - b\sqrt {n})^4 = 8$ produces $a^4 + 6a^2b^2n + b^4n^2 = 4$ . The change of variable $b^2n = c- 3a^2$ now yields $c^2 - 8a^4 = 4$ . Using the rational point $(a, c) = (0,2)$ and [Reference Mordell9, Theorem 2 on page 77] yields a birational equivalence with the elliptic curve $y^2 = x^3 - 8x$ LMFDB label 256.c1, which has Mordell–Weil group $\mathbb {Z} / 2\mathbb {Z}$ and only the trivial rational point $(0, 0)$ . Working backwards from this point and the point at infinity, we obtain two points $(x:y:z) = (0:2:1)$ and $(0:1:0)$ on the projective version of our curve $y^2z^2-8x^4=4z^4$ . The only rational point on $c^2 - 8a^4 = 4$ is $(a, c) = (0,2)$ . Since $a=0$ , this contradicts our assumption that a and b are both nonzero.