[proofplan] We analyse the reduction of the Frey curve prime by prime using the Weierstrass invariants of its defining equation. At odd primes, the discriminant and $c_4$ invariant show that primes not dividing $abc$ give good reduction, while primes dividing $abc$ give multiplicative reduction. The prime $2$ is handled by passing to the standard integral model adapted to the parity normalization $b$ even and $a \equiv -1 \pmod 4$; in that model $c_4$ is a $2$-adic unit and the discriminant has positive valuation, so the reduction is multiplicative. These alternatives exhaust all rational primes, proving semistability. [/proofplan] [step:Record the invariants of the displayed Frey model] Define the integers \begin{align*} A := a^p,\qquad B := b^p,\qquad C := -c^p. \end{align*} Then the Fermat relation gives $A+B=C$. The displayed model is \begin{align*} E_{a,b,p}: \quad y^2 = x(x-A)(x+B) = x^3 + (B-A)x^2 - ABx. \end{align*} For the integral Weierstrass equation \begin{align*} y^2 = x^3 + a_2x^2 + a_4x + a_6 \end{align*} with $a_2=B-A$, $a_4=-AB$, and $a_6=0$, the standard Weierstrass invariants are \begin{align*} b_2 &= 4(B-A),\\ b_4 &= -2AB,\\ b_6 &= 0,\\ c_4 &= b_2^2 - 24b_4 = 16(A^2+AB+B^2),\\ \Delta &= 16A^2B^2(A+B)^2 = 16(abc)^{2p}. \end{align*} Here the formula for $\Delta$ is the discriminant formula for the cubic with roots $0$, $A$, and $-B$: the pairwise root differences are $A$, $-B$, and $A+B=C$. For every rational prime $\ell$, let $v_\ell: \mathbb{Q}^\times \to \mathbb{Z}$ denote the normalized $\ell$-adic valuation, characterized by $v_\ell(\ell)=1$ and $v_\ell(u)=0$ for every rational number $u$ whose numerator and denominator are not divisible by $\ell$; set $v_\ell(0):=\infty$. [guided] We first put the curve into a form where the local computations can be read directly from the coefficients. Define \begin{align*} A := a^p,\qquad B := b^p,\qquad C := -c^p. \end{align*} Since $a^p+b^p+c^p=0$, we have $A+B=C$. The Frey curve is therefore \begin{align*} E_{a,b,p}: \quad y^2 = x(x-A)(x+B). \end{align*} Expanding the right-hand side gives \begin{align*} x(x-A)(x+B) = x(x^2+(B-A)x-AB) = x^3+(B-A)x^2-ABx. \end{align*} Thus this is a Weierstrass equation of the form \begin{align*} y^2 = x^3 + a_2x^2 + a_4x + a_6 \end{align*} with \begin{align*} a_2=B-A,\qquad a_4=-AB,\qquad a_6=0. \end{align*} For this form the auxiliary invariants are \begin{align*} b_2 &= 4a_2 = 4(B-A),\\ b_4 &= 2a_4 = -2AB,\\ b_6 &= 4a_6 = 0. \end{align*} Therefore \begin{align*} c_4 &= b_2^2 - 24b_4 \\ &= 16(B-A)^2 + 48AB \\ &= 16(A^2 - 2AB + B^2 + 3AB) \\ &= 16(A^2+AB+B^2). \end{align*} The discriminant of the cubic $x(x-A)(x+B)$ is the square of the product of the pairwise differences of the roots $0$, $A$, and $-B$. These differences are $A$, $-B$, and $A+B=C$. Hence the Weierstrass discriminant is \begin{align*} \Delta = 16A^2B^2(A+B)^2 = 16A^2B^2C^2 = 16(abc)^{2p}. \end{align*} This is the key numerical input for the local reduction analysis. For every rational prime $\ell$, we write $v_\ell: \mathbb{Q}^\times \to \mathbb{Z}$ for the normalized $\ell$-adic valuation. Thus $v_\ell(\ell)=1$, and $v_\ell(u)=0$ whenever $u$ is a rational number whose numerator and denominator are both prime to $\ell$. We also set $v_\ell(0):=\infty$. This notation will be used to distinguish unit invariants from invariants divisible by the prime under consideration. [/guided] [/step] [step:Show good reduction at odd primes not dividing $abc$] Let $\ell$ be an odd rational prime with $\ell \nmid abc$. Then $\ell \nmid \Delta$ because \begin{align*} \Delta = 16(abc)^{2p} \end{align*} and $\ell \neq 2$. Hence the displayed integral Weierstrass equation has discriminant an $\ell$-adic unit. Therefore $E_{a,b,p}$ has good reduction at $\ell$. [/step] [step:Show multiplicative reduction at odd primes dividing $abc$] Let $\ell$ be an odd rational prime with $\ell \mid abc$. Since $\gcd(a,b,c)=1$ and $a^p+b^p+c^p=0$, exactly one of $a,b,c$ is divisible by $\ell$. We have $\ell \mid \Delta$, so the curve has bad reduction at $\ell$ in the displayed integral model. We now check that $c_4$ is an $\ell$-adic unit. If $\ell \mid a$, then $A \equiv 0 \pmod{\ell}$ while $B \not\equiv 0 \pmod{\ell}$, so \begin{align*} A^2+AB+B^2 \equiv B^2 \not\equiv 0 \pmod{\ell}. \end{align*} If $\ell \mid b$, then \begin{align*} A^2+AB+B^2 \equiv A^2 \not\equiv 0 \pmod{\ell}. \end{align*} If $\ell \mid c$, then $A+B=C \equiv 0 \pmod{\ell}$, so $B \equiv -A \pmod{\ell}$ and \begin{align*} A^2+AB+B^2 \equiv A^2-A^2+A^2 = A^2 \not\equiv 0 \pmod{\ell}. \end{align*} Because $\ell$ is odd, $16$ is also an $\ell$-adic unit. Hence $v_\ell(c_4)=0$. We use the following local consequence of Tate's algorithm: for a minimal integral Weierstrass model over $\mathbb{Q}_\ell$, if $v_\ell(\Delta)>0$ and $v_\ell(c_4)=0$, then the reduction is multiplicative (citing a result not yet in the wiki: Tate's algorithm multiplicative reduction criterion). The condition $v_\ell(c_4)=0$ is invariant under passing from the displayed integral model to a minimal integral model at $\ell$, and the $j$-invariant satisfies \begin{align*} v_\ell(j(E_{a,b,p})) = v_\ell(c_4^3)-v_\ell(\Delta) = -v_\ell(\Delta) <0. \end{align*} Thus the reduction at $\ell$ is multiplicative. [guided] Now consider an odd prime $\ell$ which divides $abc$. The first point is that primitivity forces only one of the three integers to be divisible by $\ell$. Indeed, if $\ell$ divided two of $a,b,c$, then the Fermat relation \begin{align*} a^p+b^p+c^p=0 \end{align*} would force $\ell$ to divide the third as well, contradicting $\gcd(a,b,c)=1$. Since \begin{align*} \Delta = 16(abc)^{2p}, \end{align*} we have $\ell \mid \Delta$. Thus the displayed integral model is singular modulo $\ell$. To decide whether this bad reduction is multiplicative or additive, we inspect $c_4$. The invariant is \begin{align*} c_4 = 16(A^2+AB+B^2). \end{align*} Because $\ell$ is odd, the factor $16$ is invertible modulo $\ell$, so it is enough to prove \begin{align*} A^2+AB+B^2 \not\equiv 0 \pmod{\ell}. \end{align*} There are three cases, corresponding to which of $a,b,c$ is divisible by $\ell$. If $\ell \mid a$, then $A=a^p \equiv 0 \pmod{\ell}$, while $B=b^p \not\equiv 0 \pmod{\ell}$. Therefore \begin{align*} A^2+AB+B^2 \equiv B^2 \not\equiv 0 \pmod{\ell}. \end{align*} If $\ell \mid b$, then $B \equiv 0 \pmod{\ell}$ and $A \not\equiv 0 \pmod{\ell}$, so \begin{align*} A^2+AB+B^2 \equiv A^2 \not\equiv 0 \pmod{\ell}. \end{align*} If $\ell \mid c$, then $C=-c^p \equiv 0 \pmod{\ell}$. Since $A+B=C$, we get $B \equiv -A \pmod{\ell}$. Also $A \not\equiv 0 \pmod{\ell}$, because otherwise $\ell$ would divide both $a$ and $c$. Hence \begin{align*} A^2+AB+B^2 \equiv A^2 + A(-A) + (-A)^2 = A^2 \not\equiv 0 \pmod{\ell}. \end{align*} Thus $v_\ell(c_4)=0$, while $v_\ell(\Delta)>0$. The local reduction criterion supplied by Tate's algorithm says: for a minimal integral Weierstrass model over $\mathbb{Q}_\ell$, the conditions $v_\ell(\Delta)>0$ and $v_\ell(c_4)=0$ imply multiplicative reduction (citing a result not yet in the wiki: Tate's algorithm multiplicative reduction criterion). Passing to a minimal model does not change the fact that the $j$-invariant has negative valuation, since \begin{align*} v_\ell(j(E_{a,b,p})) = v_\ell(c_4^3)-v_\ell(\Delta) = -v_\ell(\Delta) <0. \end{align*} Therefore the reduction at every odd prime dividing $abc$ is multiplicative. [/guided] [/step] [step:Replace the equation at $2$ by a standard integral model adapted to the parity normalization] Because the triple is primitive and $p$ is odd, exactly one of $a,b,c$ is even. By the chosen normalization, $b$ is even and $a,c$ are odd, with $a \equiv -1 \pmod 4$. Since $p \geq 5$, the integer $B=b^p$ is divisible by $32$. Define the integral Weierstrass model \begin{align*} E' : \quad Y^2 + XY = X^3 + \frac{B-A-1}{4}X^2 - \frac{AB}{16}X. \end{align*} The coefficients are integral: $AB/16 \in \mathbb{Z}$ because $16 \mid B$, and \begin{align*} B-A-1 \equiv 0-(-1)-1 \equiv 0 \pmod 4 \end{align*} because $B \equiv 0 \pmod 4$ and $A=a^p \equiv a \equiv -1 \pmod 4$. This model is obtained from the displayed model by the change of variables \begin{align*} x = 4X,\qquad y = 8Y+4X. \end{align*} Indeed, substituting these expressions into \begin{align*} y^2=x(x-A)(x+B) \end{align*} and dividing by $16$ gives \begin{align*} (2Y+X)^2 = X(4X-A)(4X+B), \end{align*} which is equivalent to \begin{align*} Y^2 + XY = X^3 + \frac{B-A-1}{4}X^2 - \frac{AB}{16}X. \end{align*} For this model the invariants are \begin{align*} c_4' &= A^2+AB+B^2,\\ \Delta' &= \frac{A^2B^2C^2}{256}. \end{align*} Since $A$ and $C$ are odd and $B$ is even, \begin{align*} c_4' \equiv A^2 \equiv 1 \pmod 2, \end{align*} so $v_2(c_4')=0$. Also \begin{align*} v_2(\Delta') = 2v_2(B)-8 = 2p\,v_2(b)-8 \geq 2p-8 \geq 2, \end{align*} so $v_2(\Delta')>0$. By the same Tate-algorithm multiplicative reduction criterion, the reduction at $2$ is multiplicative. [guided] The prime $2$ cannot be handled by the displayed equation alone, because its coefficients have unnecessary powers of $2$ in the invariants. We use the parity normalization to write down the correct integral model. Since $p$ is odd and \begin{align*} a^p+b^p+c^p=0, \end{align*} the parities of $a^p,b^p,c^p$ are the same as the parities of $a,b,c$. Primitivity excludes two of $a,b,c$ being even. The sum of three odd integers is odd, so not all three are odd. Hence exactly one of $a,b,c$ is even. By our normalization this even integer is $b$, and $a,c$ are odd. We have also chosen the sign of $a$ so that \begin{align*} a \equiv -1 \pmod 4. \end{align*} Since $p$ is odd, this gives \begin{align*} A=a^p \equiv a \equiv -1 \pmod 4. \end{align*} Since $b$ is even and $p \geq 5$, the integer \begin{align*} B=b^p \end{align*} is divisible by $2^p$, hence by $32$. Now define \begin{align*} E' : \quad Y^2 + XY = X^3 + \frac{B-A-1}{4}X^2 - \frac{AB}{16}X. \end{align*} We verify that this is an integral Weierstrass model. First, $AB/16 \in \mathbb{Z}$ because $16 \mid B$. Second, \begin{align*} B-A-1 \equiv 0-(-1)-1 \equiv 0 \pmod 4, \end{align*} so $(B-A-1)/4 \in \mathbb{Z}$. This model is not introduced out of nowhere; it is obtained from the original equation by the change of variables \begin{align*} x = 4X,\qquad y = 8Y+4X. \end{align*} Substituting into the original equation gives \begin{align*} (8Y+4X)^2 = 4X(4X-A)(4X+B). \end{align*} Dividing by $16$ yields \begin{align*} (2Y+X)^2 = X(4X-A)(4X+B). \end{align*} Expanding both sides, \begin{align*} 4Y^2+4XY+X^2 = 16X^3+4(B-A)X^2-ABX. \end{align*} Dividing by $4$ and moving $X^2/4$ to the right gives \begin{align*} Y^2+XY = X^3+\frac{B-A-1}{4}X^2-\frac{AB}{16}X. \end{align*} For this transformed model the invariants are scaled from the original model by the substitution with scaling factor $2$. Thus \begin{align*} c_4' = \frac{c_4}{16}=A^2+AB+B^2, \qquad \Delta' = \frac{\Delta}{4096}=\frac{A^2B^2C^2}{256}. \end{align*} We now inspect their $2$-adic valuations. Since $A$ and $C$ are odd and $B$ is even, \begin{align*} c_4' = A^2+AB+B^2 \equiv A^2 \equiv 1 \pmod 2. \end{align*} Hence $v_2(c_4')=0$. For the discriminant, \begin{align*} v_2(\Delta') = v_2(A^2)+v_2(B^2)+v_2(C^2)-8 = 2v_2(B)-8 = 2p\,v_2(b)-8. \end{align*} Because $b$ is even, $v_2(b)\geq 1$, and because $p\geq 5$, \begin{align*} v_2(\Delta') \geq 2p-8 \geq 2. \end{align*} Thus $v_2(\Delta')>0$ while $v_2(c_4')=0$. The Tate-algorithm multiplicative reduction criterion therefore gives multiplicative reduction at $2$. [/guided] [/step] [step:Conclude semistability at every rational prime] Every rational prime $\ell$ belongs to exactly one of the following cases: $\ell=2$, $\ell$ is odd and $\ell \mid abc$, or $\ell$ is odd and $\ell \nmid abc$. The preceding steps prove multiplicative reduction in the first two cases and good reduction in the third case. Hence every minimal Weierstrass model of $E_{a,b,p}$ has only good or multiplicative reduction at every rational prime. Therefore $E_{a,b,p}$ is semistable over $\mathbb{Q}$. [/step]