[proofplan] We compute the discriminant and the invariant $c_4$ of the given integral Weierstrass model. At an odd prime not dividing $abc$, the discriminant is a unit, so the curve has good reduction and conductor exponent $0$. At an odd prime dividing $abc$, primitivity forces exactly one of $a,b,c$ to vanish modulo $\ell$; the discriminant has positive valuation while $c_4$ is a unit, giving multiplicative reduction and hence conductor exponent $1$. The prime $2$ is excluded throughout because the statement concerns only the odd part of the conductor. [/proofplan] [step:Compute the discriminant and $c_4$ of the Frey model] Define integers \begin{align*} A := a^p, \qquad B := b^p. \end{align*} The Frey curve is given by the integral Weierstrass equation \begin{align*} y^2 = x^3 + (B-A)x^2 - ABx. \end{align*} For this model, the standard Weierstrass quantities are \begin{align*} b_2 &= 4(B-A),\\ b_4 &= -2AB,\\ b_6 &= 0. \end{align*} Therefore \begin{align*} c_4 &= b_2^2 - 24b_4 \\ &= 16(B-A)^2 + 48AB \\ &= 16(A^2 + AB + B^2). \end{align*} The discriminant of the cubic with roots $0$, $A$, and $-B$ is \begin{align*} A^2B^2(A+B)^2, \end{align*} and the Weierstrass discriminant is therefore \begin{align*} \Delta = 16A^2B^2(A+B)^2. \end{align*} Since $A+B=a^p+b^p=c^p$, this becomes \begin{align*} \Delta = 16(a b c)^{2p}. \end{align*} [guided] We first put the curve into the standard cubic form. With \begin{align*} A := a^p, \qquad B := b^p, \end{align*} the equation \begin{align*} y^2 = x(x-A)(x+B) \end{align*} expands to \begin{align*} y^2 = x^3 + (B-A)x^2 - ABx. \end{align*} For a Weierstrass equation of the form \begin{align*} y^2 = x^3 + a_2x^2 + a_4x + a_6, \end{align*} with $a_1=a_3=0$, the quantities used to test reduction are \begin{align*} b_2 = 4a_2, \qquad b_4 = 2a_4, \qquad b_6 = 4a_6. \end{align*} Here $a_2=B-A$, $a_4=-AB$, and $a_6=0$, so \begin{align*} b_2 &= 4(B-A),\\ b_4 &= -2AB,\\ b_6 &= 0. \end{align*} Thus \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 is equally explicit because the cubic factors completely: \begin{align*} x(x-A)(x+B). \end{align*} Its roots are $0$, $A$, and $-B$. The square of the product of pairwise differences is \begin{align*} (0-A)^2(0+B)^2(A+B)^2 = A^2B^2(A+B)^2. \end{align*} For this Weierstrass model, the elliptic discriminant is $16$ times that cubic discriminant, hence \begin{align*} \Delta = 16A^2B^2(A+B)^2. \end{align*} Finally, the Fermat equation gives $A+B=a^p+b^p=c^p$, so \begin{align*} \Delta = 16(a b c)^{2p}. \end{align*} [/guided] [/step] [step:Show that odd primes not dividing $abc$ have good reduction] Let $\ell$ be an odd prime such that $\ell \nmid abc$. Since $\ell \neq 2$, the integer $16$ is a unit modulo $\ell$. Since $\ell \nmid abc$, the discriminant \begin{align*} \Delta = 16(abc)^{2p} \end{align*} satisfies $v_\ell(\Delta)=0$, where $v_\ell: \mathbb{Q}^{\times} \to \mathbb{Z}$ is the $\ell$-adic valuation. Hence the reduction of the displayed integral Weierstrass model modulo $\ell$ is nonsingular, so $E_{a,b,p}$ has good reduction at $\ell$. By the local definition of the conductor exponent for an elliptic curve, good reduction gives \begin{align*} f_\ell(E_{a,b,p}) = 0. \end{align*} [/step] [step:Show that odd primes dividing $abc$ have multiplicative reduction] Let $\ell$ be an odd prime such that $\ell \mid abc$. Since $a,b,c$ are pairwise coprime, exactly one of $a$, $b$, and $c$ is divisible by $\ell$. If $\ell \mid a$, then $A \equiv 0 \pmod{\ell}$ while $B \not\equiv 0 \pmod{\ell}$. Hence \begin{align*} A^2 + AB + B^2 \equiv B^2 \not\equiv 0 \pmod{\ell}. \end{align*} If $\ell \mid b$, then $B \equiv 0 \pmod{\ell}$ while $A \not\equiv 0 \pmod{\ell}$, so \begin{align*} A^2 + AB + B^2 \equiv A^2 \not\equiv 0 \pmod{\ell}. \end{align*} If $\ell \mid c$, then $A+B=c^p \equiv 0 \pmod{\ell}$, so $B \equiv -A \pmod{\ell}$. Since $\ell \nmid a$, we have $A \not\equiv 0 \pmod{\ell}$, and therefore \begin{align*} A^2 + AB + B^2 \equiv A^2 - A^2 + A^2 = A^2 \not\equiv 0 \pmod{\ell}. \end{align*} In all three cases, \begin{align*} v_\ell(c_4)=0. \end{align*} On the other hand, since $\ell \mid abc$, \begin{align*} v_\ell(\Delta)=v_\ell\bigl(16(abc)^{2p}\bigr)>0. \end{align*} Thus the reduction is singular while $c_4$ is an $\ell$-adic unit. By the standard local reduction criterion for elliptic curves, this is multiplicative reduction. For multiplicative reduction, the local conductor exponent is \begin{align*} f_\ell(E_{a,b,p}) = 1. \end{align*} [guided] Now suppose $\ell$ is an odd prime dividing $abc$. The primitivity hypothesis is crucial here: because $a$, $b$, and $c$ are pairwise coprime, $\ell$ can divide only one of them. We inspect $c_4=16(A^2+AB+B^2)$ modulo $\ell$. Since $\ell$ is odd, $16$ is a unit modulo $\ell$, so it is enough to check that $A^2+AB+B^2$ is nonzero modulo $\ell$. If $\ell \mid a$, then $A=a^p \equiv 0 \pmod{\ell}$, while $\ell \nmid b$ gives $B=b^p \not\equiv 0 \pmod{\ell}$. Therefore \begin{align*} A^2 + AB + B^2 \equiv B^2 \not\equiv 0 \pmod{\ell}. \end{align*} If $\ell \mid b$, the same computation with $A$ and $B$ interchanged gives \begin{align*} A^2 + AB + B^2 \equiv A^2 \not\equiv 0 \pmod{\ell}. \end{align*} If $\ell \mid c$, then the Fermat equation gives \begin{align*} A+B = c^p \equiv 0 \pmod{\ell}, \end{align*} so $B \equiv -A \pmod{\ell}$. Since $\ell$ cannot also divide $a$, we have $A \not\equiv 0 \pmod{\ell}$. Substituting $B \equiv -A$ gives \begin{align*} A^2 + AB + B^2 \equiv A^2 - A^2 + A^2 = A^2 \not\equiv 0 \pmod{\ell}. \end{align*} Thus in every possible case, \begin{align*} v_\ell(c_4)=0. \end{align*} At the same time, the discriminant is \begin{align*} \Delta = 16(abc)^{2p}. \end{align*} Because $\ell \mid abc$ and $\ell \neq 2$, this gives \begin{align*} v_\ell(\Delta)>0. \end{align*} So the reduction modulo $\ell$ is singular, but $c_4$ remains an $\ell$-adic unit. The standard local reduction criterion for elliptic curves says exactly that this combination, $v_\ell(\Delta)>0$ and $v_\ell(c_4)=0$, is multiplicative reduction. Multiplicative reduction contributes conductor exponent $1$, hence \begin{align*} f_\ell(E_{a,b,p}) = 1. \end{align*} [/guided] [/step] [step:Assemble the odd part of the conductor] For every odd prime $\ell$, the previous two steps prove \begin{align*} f_\ell(E_{a,b,p}) = \begin{cases} 1, & \ell \mid abc,\\ 0, & \ell \nmid abc. \end{cases} \end{align*} The odd part of $N(E_{a,b,p})$ is obtained by multiplying $\ell^{f_\ell(E_{a,b,p})}$ over all odd primes $\ell$. Therefore \begin{align*} \prod_{\ell \text{ odd}} \ell^{f_\ell(E_{a,b,p})} = \prod_{\substack{\ell \mid abc\\ \ell \text{ odd}}} \ell. \end{align*} This proves the stated formula for the odd part of the conductor. [/step]