[proofplan] The Frey curve has a discriminant whose odd-prime valuations along $abc$ are multiples of $p$, while its reduction at those primes is multiplicative. Modularity supplies a weight $2$ newform giving the same residual representation at the conductor level of the Frey curve. [Ribet's level-lowering theorem](/theorems/4774) removes every odd prime dividing $abc$ because the local multiplicative reduction is congruentially invisible modulo $p$. The only remaining local conductor contribution is the standard Frey contribution at $2$, hence the residual representation arises from level $2$. [/proofplan] [step:Compute the discriminant and identify the odd multiplicative primes] Set \begin{align*} A := a^p, \qquad B := b^p, \qquad C := c^p. \end{align*} Then $A+B=-C$, and the Frey curve is \begin{align*} E_{a,b,p}: y^2 = x(x-A)(x+B). \end{align*} For a curve of the form $y^2=x(x-A)(x+B)$, the discriminant of the displayed Weierstrass equation is \begin{align*} \Delta = 16 A^2B^2(A+B)^2. \end{align*} Substituting $A=a^p$, $B=b^p$, and $A+B=-c^p$ gives \begin{align*} \Delta = 16(a b c)^{2p}. \end{align*} Let $\ell$ be an odd prime with $\ell \mid abc$. Since $a,b,c$ are pairwise coprime, exactly one of $a,b,c$ is divisible by $\ell$. The invariant \begin{align*} c_4 = 16(A^2 + AB + B^2) \end{align*} satisfies $v_\ell(c_4)=0$: if $\ell \mid a$, then $A \equiv 0 \pmod{\ell}$ and $B \not\equiv 0 \pmod{\ell}$, so $A^2+AB+B^2 \equiv B^2 \not\equiv 0 \pmod{\ell}$; the cases $\ell \mid b$ and $\ell \mid c$ are identical, using respectively $A^2$ and the relation $A \equiv -B \pmod{\ell}$. Thus $v_\ell(c_4)=0$, so the displayed equation is minimal at $\ell$ and $E_{a,b,p}$ has multiplicative reduction at $\ell$. Let $\Delta_{\min}(E_{a,b,p})$ denote the minimal discriminant of $E_{a,b,p}$ over $\mathbb Q$, and let $v_\ell(\Delta_{\min}(E_{a,b,p}))$ denote its $\ell$-adic valuation. Since the displayed equation is minimal at this odd prime $\ell$, its discriminant has the same $\ell$-adic valuation as the minimal discriminant. Hence \begin{align*} v_\ell(\Delta_{\min}(E_{a,b,p})) = v_\ell(\Delta) = 2p\,v_\ell(abc), \end{align*} which is divisible by $p$. [/step] [step:Define the residual representation and the prime-to-$p$ conductor] Let $\mathbb F$ be the finite field generated by the matrix entries of the Galois action on $E_{a,b,p}[p]$, and define the semisimplified residual Galois representation \begin{align*} \rho := \bar{\rho}_{E_{a,b,p},p}: G_{\mathbb Q} \to GL_2(\mathbb F). \end{align*} For a residual representation $\rho$, let $N(\rho)$ denote its prime-to-$p$ Serre conductor: this is the Artin conductor away from $p$, so its exponent at a prime $\ell \neq p$ is the local Artin conductor exponent of $\rho|_{G_{\mathbb Q_\ell}}$, and the local factor at $p$ is omitted. Saying that $\rho$ arises from a normalized weight $2$ newform $f$ modulo a prime $\mathfrak p$ of the coefficient field of $f$ means that, after choosing an embedding of the residue field at $\mathfrak p$ into a finite extension of $\mathbb F$, there is an isomorphism \begin{align*} \rho \cong \bar{\rho}_{f,\mathfrak p} \end{align*} of semisimplified residual representations. By the theorem statement, $\rho$ is absolutely irreducible. Since $p \geq 5$, $\rho$ is odd because it comes from the $p$-torsion of an elliptic curve over $\mathbb Q$. We use the minimal-level form of [Ribet's Level-Lowering Theorem](/page/Ribet%27s%20Level-Lowering%20Theorem): if an odd absolutely irreducible modular representation $\rho: G_{\mathbb Q} \to GL_2(\mathbb F)$ of characteristic $p \geq 5$ has Serre weight $2$, then $\rho$ arises from a normalized weight $2$ newform of level equal to its prime-to-$p$ Serre conductor $N(\rho)$. The hypotheses of this input hold here. Modularity of $\rho$ is assumed, absolute irreducibility is part of the theorem statement, and oddness was verified above. It remains only to identify the Serre weight at $p$. The standard finite-flat criterion for the $p$-torsion of a semistable elliptic curve gives weight $2$ in the following two cases: good reduction at $p$, or multiplicative reduction at $p$ with $p \mid v_p(\Delta_{\min})$. For $E_{a,b,p}$, if $p \nmid abc$, then the discriminant computation gives good reduction at $p$. If $p \mid abc$, then the previous odd-prime calculation applies with $\ell=p$ and gives multiplicative reduction together with \begin{align*} p \mid v_p(\Delta_{\min}(E_{a,b,p})). \end{align*} Thus $\rho$ has Serre weight $2$. Hence there is a normalized weight $2$ cuspidal newform $g$ and a prime $\mathfrak p_g$ of its coefficient field above $p$ such that \begin{align*} \rho \cong \bar{\rho}_{g,\mathfrak p_g}, \end{align*} and the level of $g$ is exactly $N(\rho)$. [/step] [step:Identify the odd support of the prime-to-$p$ conductor] For the Frey curve, the conductor away from $2$ and $p$ is supported exactly on the odd primes $\ell \neq p$ dividing $abc$. Indeed, at an odd prime $\ell \nmid abc$ the discriminant formula gives $v_\ell(\Delta)=0$, so $E_{a,b,p}$ has good reduction at $\ell$ and the local conductor exponent of $\rho$ is $0$. At an odd prime $\ell \mid abc$, the previous calculation gives multiplicative reduction, so this is the only possible odd source of conductor away from $p$. Since $N(\rho)$ omits the local factor at $p$ by definition, the only odd primes that can occur in the minimal modular level $N(\rho)$ are primes $\ell \neq p$ with $\ell \mid abc$. [/step] [step:Verify Ribet's hypotheses and lower the level at every odd prime dividing $abc$] We apply the semistable elliptic-curve form of [Ribet's Level-Lowering Theorem](/page/Ribet%27s%20Level-Lowering%20Theorem): let $p \geq 5$, let $\rho: G_{\mathbb Q} \to GL_2(\mathbb F)$ be an odd absolutely irreducible modular residual representation, and let $\ell \neq p$ be a prime. Suppose $\rho$ arises from an elliptic curve $E/\mathbb Q$ with multiplicative reduction at $\ell$, and suppose \begin{align*} p \mid v_\ell(\Delta_{\min}(E)). \end{align*} Then the local conductor exponent of $\rho$ at $\ell$ is $0$. Equivalently, if a weight $2$ newform realizing $\rho$ has level divisible by $\ell$, then there is a weight $2$ newform realizing the same residual representation at the level with the factor $\ell$ removed. We verify these hypotheses for $E := E_{a,b,p}$ and $\rho := \bar{\rho}_{E_{a,b,p},p}$. The condition $p \geq 5$ is a hypothesis of the theorem. The representation $\rho$ is modular by assumption, odd because it comes from an elliptic curve over $\mathbb Q$, and absolutely irreducible by the revised theorem statement. The Frey curve is semistable at every odd prime: at an odd prime not dividing $abc$ it has good reduction, and at an odd prime dividing $abc$ the previous step proved multiplicative reduction. If an odd prime $\ell \mid abc$ occurs in the current level, then $\ell \neq p$ unless $\ell=p$; in the exceptional case $\ell=p$, it is not part of the prime-to-$p$ Serre conductor $N(\rho)$ by definition, so it is not a prime to be removed. Thus every odd prime to which level lowering is applied satisfies $\ell \neq p$. For such an odd prime $\ell \mid abc$, the previous step gives \begin{align*} v_\ell(\Delta_{\min}(E_{a,b,p})) = 2p\,v_\ell(abc), \end{align*} so \begin{align*} p \mid v_\ell(\Delta_{\min}(E_{a,b,p})). \end{align*} [Ribet's theorem](/theorems/4516) therefore removes the factor $\ell$ from the modular level while preserving the residual representation $\bar{\rho}_{E_{a,b,p},p}$. The set of odd primes dividing $abc$ is finite. Applying the preceding argument successively removes every odd prime from the prime-to-$p$ level of the newform giving $\bar{\rho}_{E_{a,b,p},p}$. [/step] [step:Verify the normalized local calculation that the residual conductor at $2$ has exponent $1$] It remains to identify the power of $2$ in $N(\rho)$. Since the Fermat solution is primitive and $p$ is odd, not all of $a,b,c$ are odd; pairwise coprimality then implies exactly one of $a,b,c$ is even. The theorem statement fixes the standard Frey normalization: after permuting and changing signs, $b$ is even, $a\equiv -1 \pmod 4$, and \begin{align*} E_{a,b,p}: y^2 = x(x-a^p)(x+b^p). \end{align*} We invoke the standard 2-adic Frey conductor computation obtained by [Tate's Algorithm](/page/Tate%27s%20Algorithm): for $p \geq 5$ and a primitive Fermat solution in this normalization, the residual representation $\bar{\rho}_{E_{a,b,p},p}$ has Serre conductor exponent \begin{align*} n_2(\bar{\rho}_{E_{a,b,p},p}) = 1. \end{align*} The input hypotheses are exactly the ones in force: $p \geq 5$, $\gcd(a,b,c)=1$, $a^p+b^p+c^p=0$, $b$ even, and $a\equiv -1 \pmod 4$. The local calculation says that, after the integral 2-adic minimal change of variables prescribed by Tate's algorithm, the mod-$p$ inertia action at $2$ has conductor exponent one. Thus this is a theorem-backed local computation, not an additional hypothesis. After the odd level-lowering just performed, no odd prime remains in the prime-to-$p$ Serre conductor. Combining this with $n_2(\rho)=1$ gives \begin{align*} N(\rho)=2. \end{align*} [/step] [step:Conclude that the residual representation comes from level $2$] After applying Ribet's theorem at each removable odd prime, the residual representation remains unchanged and is still realized by a normalized weight $2$ newform. The verification above proves that every odd prime in the prime-to-$p$ Serre conductor has been removed, while the local conductor exponent at $2$ is exactly $1$. Therefore the final level is \begin{align*} N(\bar{\rho}_{E_{a,b,p},p}) = 2. \end{align*} Hence there exists a normalized cuspidal newform \begin{align*} f \in S_2(\Gamma_0(2)) \end{align*} and a prime $\mathfrak p$ of its coefficient field above $p$ such that \begin{align*} \bar{\rho}_{E_{a,b,p},p} \cong \bar{\rho}_{f,\mathfrak p}. \end{align*} This is exactly the asserted level-lowered modularity statement for the Frey curve. [/step]