[proofplan] We argue by contradiction from a primitive first-case Fermat solution. Reducing the equation modulo the auxiliary prime $q$ shows that $q$ must divide one of the three variables, because otherwise two nonzero $p$th-power residues modulo $q$ would differ by $1$. The Barlow-Abel auxiliary-prime lemma then converts this forced divisibility into the assertion that $p$ is itself a $p$th-power residue modulo $q$, contradicting the second hypothesis. [/proofplan] [step:Assume a primitive first-case Fermat solution exists] Suppose, toward a contradiction, that the first case fails. Then there exist nonzero integers $a,b,c \in \mathbb{Z}$ such that \begin{align*} a^p+b^p+c^p=0 \end{align*} and $p \nmid abc$. Dividing by the common divisor $\gcd(a,b,c)$ if necessary, we may also assume $\gcd(a,b,c)=1$; this preserves the displayed equation and the condition $p \nmid abc$. [guided] To disprove the existence of a first-case Fermat solution, we assume one exists and derive a contradiction. Thus we take nonzero integers $a,b,c \in \mathbb{Z}$ satisfying \begin{align*} a^p+b^p+c^p=0 \end{align*} with $p \nmid abc$. The phrase "first case" means exactly that the exponent $p$ divides none of $a,b,c$. We may replace $(a,b,c)$ by the primitive triple obtained after dividing by $\gcd(a,b,c)$, because Fermat's equation is homogeneous of degree $p$ and the divisibility condition $p \nmid abc$ remains true after removing a common divisor. Hence we assume $\gcd(a,b,c)=1$. [/guided] [/step] [step:Force the auxiliary prime $q$ to divide one variable] We prove that $q \mid abc$. Suppose instead that $q \nmid abc$. Since $q$ is prime, the residue classes of $a,b,c$ are nonzero in $\mathbb{F}_q$. Define \begin{align*} R_q:=\{u^p \in \mathbb{F}_q^\times : u \in \mathbb{F}_q^\times\} \end{align*} to be the set of nonzero $p$th-power residues modulo $q$. Because $c \not\equiv 0 \pmod q$, the elements \begin{align*} r:=(-a c^{-1})^p \in R_q, \qquad s:=(b c^{-1})^p \in R_q \end{align*} are well-defined and nonzero in $\mathbb{F}_q$. Since $p$ is odd, $(-a)^p=-a^p$, and the Fermat equation gives \begin{align*} r-s &=(-a c^{-1})^p-(b c^{-1})^p \\ &=\frac{-a^p-b^p}{c^p} \\ &=\frac{c^p}{c^p} \\ &=1 \end{align*} in $\mathbb{F}_q$. Thus two nonzero $p$th-power residues modulo $q$ differ by $1$, contradicting hypothesis 1. Therefore $q \mid abc$. [guided] The first residue condition is designed to rule out a Fermat solution after reducing modulo $q$. Assume for contradiction within this step that $q$ divides none of $a,b,c$. Since $q$ is prime, each of $a,b,c$ then has a multiplicative inverse modulo $q$. Let \begin{align*} R_q:=\{u^p \in \mathbb{F}_q^\times : u \in \mathbb{F}_q^\times\} \end{align*} be the set of nonzero $p$th-power residues in the finite field $\mathbb{F}_q$. We form two elements of this set by normalising the Fermat equation by $c^p$: \begin{align*} r:=(-a c^{-1})^p, \qquad s:=(b c^{-1})^p. \end{align*} Both are nonzero $p$th-power residues because $a,b,c$ are nonzero modulo $q$. Since $p$ is odd, $(-a)^p=-a^p$. Therefore, in $\mathbb{F}_q$, \begin{align*} r-s &=(-a c^{-1})^p-(b c^{-1})^p \\ &=\frac{-a^p-b^p}{c^p} \\ &=\frac{c^p}{c^p} \\ &=1, \end{align*} where the third equality uses $a^p+b^p+c^p=0$. This produces two nonzero $p$th-power residues differing by $1$, exactly what hypothesis 1 forbids. Hence the assumption $q \nmid abc$ is impossible, and $q \mid abc$. [/guided] [/step] [step:Invoke the Barlow-Abel auxiliary-prime lemma] We use the classical Barlow-Abel auxiliary-prime lemma in the following precise form: if $p$ is an odd prime, $q=2kp+1$ is prime, $(a,b,c)$ is a primitive first-case solution of $a^p+b^p+c^p=0$, and $q$ divides one of $a,b,c$, then $p$ is a $p$th-power residue modulo $q$. The hypotheses of this lemma are satisfied: $p$ is an odd prime by the theorem statement, $q=2kp+1$ is prime by hypothesis, $(a,b,c)$ is primitive and first-case by construction, and $q \mid abc$ by the previous step. Hence $p$ is a $p$th-power residue modulo $q$. [guided] The previous step only tells us that the auxiliary prime $q$ divides one of the entries of the primitive Fermat triple. The classical Barlow-Abel auxiliary-prime lemma is the arithmetic mechanism that turns this divisibility into a residue conclusion. In the form needed here, it states: If $p$ is an odd prime, $q=2kp+1$ is prime, $(a,b,c)$ is a primitive solution of \begin{align*} a^p+b^p+c^p=0 \end{align*} with $p \nmid abc$, and $q$ divides one of $a,b,c$, then $p$ is a $p$th-power residue modulo $q$. We verify each hypothesis. The theorem assumes that $p$ is an odd prime and that $q=2kp+1$ is prime. The contradiction setup supplied a primitive solution $(a,b,c)$ with $p \nmid abc$. The preceding step proved $q \mid abc$, so $q$ divides at least one of $a,b,c$. Therefore the Barlow-Abel lemma applies and yields that $p$ is a $p$th-power residue modulo $q$. [/guided] [/step] [step:Contradict the nonresidue hypothesis and conclude] Hypothesis 2 says that $p$ is not a $p$th-power residue modulo $q$. The previous step says that $p$ is a $p$th-power residue modulo $q$. This contradiction shows that no primitive first-case Fermat solution exists. Therefore the first case of [Fermat's Last Theorem](/theorems/4789) holds for the exponent $p$. [/step]