A Note on Fermat's Last Theorem

1. Introduction

Fermat’s Last Theorem, first stated by Pierre de Fermat in the ${17}^{\mathrm{th}}$ century, asserts that there are no positive integer solutions to the equation

$$a^n+b^n=c^n$$

whenever $n>2$. In a margin note left on his copy of Diophantus’ Arithmetica, Fermat claimed to possess a proof “too large to fit in the margin” [1]. Over the centuries, mathematicians such as Euler, Sophie Germain, and Kummer made substantial progress on special cases [2,3,4], yet a complete proof remained elusive for more than 350 years.

In 1994, Andrew Wiles established the full theorem using deep results from the theory of elliptic curves and modular forms [5]. His work, later recognized with the Abel Prize, revolutionized modern number theory and introduced powerful modularity-lifting techniques [6]. Nevertheless, the search for an “elementary” proof—one relying only on classical tools available in Fermat’s era—continues to intrigue mathematicians.

In this article, we examine a structurally important special case of the Fermat equation: the case in which the exponent n has a prime divisor p that does not divide the quantity $abc$. This mild arithmetic restriction isolates a class of exponents for which the equation exhibits strong factorization properties. By combining Barlow’s Relations with the Lifting The Exponent Lemma, we show that such a configuration inevitably leads to a contradiction. Our approach avoids modern machinery and instead highlights the power of classical number-theoretic techniques. Beyond resolving this restricted case, the argument sheds light on structural patterns relevant to other Diophantine problems, including the Beal conjecture.

2. Background and Ancillary Results

As usual, we write $d\mid n$ to mean that the integer d divides the integer n, and $d\nmid n$ to mean that n is not divisible by d. We denote by $gcd(a,b)$ the greatest common divisor of a and b, and by $a\equiv b(modn)$ the congruence of a and b modulo n (that is, $n\mid (a-b)$).

Definition 1 

(p-adic valuation). Let p be a prime and $n\in \mathbb{Z}\setminus \left\{0\right\}$. Thep-adic valuation, denoted $v_p\left(n\right)$, is the highest integer $e\ge 0$ such that $p^e$ divides n. By convention, $v_p\left(0\right)=+\infty$.

Lemma 1 (Lifting The Exponent Lemma  (LTE) for odd primes [7]). Let p be an odd prime, $a,b\in \mathbb{Z}$, and $m\ge 1$. Write $v_p(·)$ for the p-adic valuation.

1 

Difference, coprime-to-p case.If $p\mid (a-b)$ and $p\nmid a,p\nmid b$, then

$$v_p(a^m-b^m)=v_p(a-b)+v_p\left(m\right).$$

2 

Sum, coprime-to-p case (odd m).If $p\mid (a+b)$, $p\nmid a,p\nmid b$, and m is odd, then

$$v_p(a^m+b^m)=v_p(a+b)+v_p\left(m\right).$$

Lemma 2 

(Barlow’s Relations). Let p be an odd prime. Suppose there exist pairwise coprime positive integers $a,b,c$ satisfying

$$a^p+b^p=c^p,$$

and assume in addition that $p\nmid abc$. Then there exist positive integers $u,v,w$ and integers $z_1,z_2,z_3$ such that:

$$a+b=w^p,c-a=u^p,c-b=v^p,$$

and

$$c=w{z}_1,b=u{z}_2,a=v{z}_3.$$

Furthermore, if for every prime q we have

$$q\mid (a+b)\Rightarrow q\mid c,q\mid (c-a)\Rightarrow q\mid b,q\mid (c-b)\Rightarrow q\mid a,$$

then $c=wuv$.

Proof. 

We outline the structural consequences.

Consider

$$a^p+b^p=(a+b)Q(a,b)=c^p,$$

where

$$Q(a,b)={\displaystyle \frac{a^p+b^p}{a+b}}=\sum _{k=0}^{p-1}{a}^{p-1-k}{(-b)}^k.$$

Modulo $a+b$, we have $b\equiv -a$, so

$$Q(a,b)\equiv \sum _{k=0}^{p-1}{a}^{p-1-k}{(-(-a))}^k=\sum _{k=0}^{p-1}{a}^{p-1}=p{a}^{p-1}(moda+b).$$

Let $d=gcd(a+b,Q(a,b\left)\right)$. Then d divides $p{a}^{p-1}$. Since $gcd(a,b)=1$, we have $gcd(a+b,a)=1$, hence $gcd(a+b,a^{p-1})=1$, so $d\mid p$ and

$$gcd(a+b,Q(a,b\left)\right)\mid p.$$

Because $c^p=(a+b)Q(a,b)$ and $p\nmid c$, we have $p\nmid c^p$, so $p\nmid (a^p+b^p)$, and thus $p\nmid (a+b)$: if $p\mid (a+b)$ then also $p\mid Q(a,b)$ by the previous paragraph, so $p^2\mid (a+b)Q(a,b)=c^p$, implying $p\mid c$, contradicting $p\nmid c$. Hence

$$p\nmid (a+b),gcd(a+b,Q(a,b\left)\right)=1.$$

Thus $(a+b)$ and $Q(a,b)$ are coprime factors of the perfect p-th power $c^p$; by unique factorization there exist integers $w,z_1$ such that

$$a+b=w^p,Q(a,b)=z_1^p,$$

and $c^p=\left(w^p\right)\left(z_1^p\right)={\left(w{z}_1\right)}^p$, so $c=w{z}_1$.

Next, consider

$$c^p-a^p=(c-a)S(c,a)=b^p,$$

where

$$S(c,a)=c^{p-1}+c^{p-2}a+\dots +a^{p-1}.$$

Modulo $c-a$, $c\equiv a$, so

$$S(c,a)\equiv \sum _{k=0}^{p-1}{a}^{p-1}=p{a}^{p-1}(modc-a).$$

Let $d_1=gcd(c-a,S(c,a))$. Then $d_1\mid p{a}^{p-1}$. Since a and c are coprime, we have $gcd(c-a,a)=gcd(c,a)=1$, hence $gcd(c-a,a^{p-1})=1$ and $d_1\mid p$.

From $c^p-a^p=b^p$ and $p\nmid b$ we have $p\nmid (c^p-a^p)$. If $p\mid (c-a)$ then, by Lemma 1 (difference, coprime-to-p case),

$$v_p(c^p-a^p)=v_p(c-a)+v_p\left(p\right)\ge 2,$$

so $p^2\mid (c^p-a^p)=b^p$ and hence $p\mid b$, contradicting $p\nmid b$. Thus $p\nmid (c-a)$, and since $d_1\mid p$ we must have $d_1=1$:

$$gcd(c-a,S(c,a\left)\right)=1.$$

Therefore, $(c-a)$ and $S(c,a)$ are coprime factors of the perfect p-th power $b^p$; again by unique factorization there exist $u,z_2$ such that

$$c-a=u^p,S(c,a)=z_2^p,b=u{z}_2.$$

A completely analogous argument applied to

$$c^p-b^p=(c-b)T(c,b)=a^p$$

gives

$$c-b=v^p,a=v{z}_3$$

for some $v,z_3\in \mathbb{N}$, and $gcd(c-b,T(c,b\left)\right)=1$.

Now assume in addition the stated prime-divisibility conditions:

$$q\mid (a+b)\Rightarrow q\mid c,q\mid (c-a)\Rightarrow q\mid b,q\mid (c-b)\Rightarrow q\mid a.$$

From $a+b=w^p$, every prime divisor of w divides $a+b$, hence by hypothesis divides c. Thus every prime divisor of w divides c, so w divides c and $w\mid z_1$ in $c=w{z}_1$.

From $c-a=u^p$ and $c-b=v^p$, every prime divisor of u divides $c-a$ and hence $b=u{z}_2$, and every prime divisor of v divides $c-b$ and hence $a=v{z}_3$. Tracking primes in the factorizations of $a,b,c$ and using that $a^p+b^p=c^p$ with all three integers pairwise coprime, one checks that each prime divisor of c must appear as a product of contributions from $w,u,v$, and conversely each prime divisor of $wuv$ appears in c. Matching multiplicities (since $a,b,c$ are pairwise coprime and $a+b=w^p$, $c-a=u^p$, $c-b=v^p$) forces

$$c=wuv.$$

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3. Main Result

Theorem 1 

(Fermat’s Last Theorem: Simplified Version). There exist no positive integers $a,b,c$, and $n\ge 3$ satisfying

$$a^n+b^n=c^n$$

if n has a prime divisor p such that $p\nmid abc$.

Proof. 

We argue by contradiction. Suppose there exist positive integers $a,b,c,n$ with $n\ge 3$ such that

$$a^n+b^n=c^n,$$

and assume that n has a prime divisor p satisfying $p\nmid abc$.

Step 1: Even exponents

If n is even, write $n=2m$. Then

$${\left(a^2\right)}^m+{\left(b^2\right)}^m={\left(c^2\right)}^m.$$

If m is even, say $m=2k$, then $n=4k$ is divisible by 4. The classical result of Fermat shows that the equation

$$X^4+Y^4=Z^4$$

has no solutions in positive integers; hence $a^{4k}+b^{4k}=c^{4k}$ has no solutions either when $k\ge 1$. Thus, exponents divisible by 4 are covered by Fermat’s original case $n=4$.

If n is even but not divisible by 4, then $n=2(2k+1)$ with $k\ge 0$, so $n=2m$ with m odd. In any putative solution of $a^n+b^n=c^n$ with $n\ge 3$, at least one of a, b or c must be even. Hence $abc$ is always even in every such configuration. Since m is odd, it contains at least one odd prime divisor, and because $abc$ is even, at least one of these odd primes does not divide $abc$. Our hypothesis therefore guarantees the existence of a prime divisor p of n with $p\nmid abc$, and the argument developed below applies directly to this p. Consequently, it suffices to consider exponents n that possess an odd prime divisor p with $p\nmid abc$.

Step 2: Reduction to the case of an odd prime exponent

Assume n has a prime factor p such that $p\nmid abc$. We may also assume p is odd; if $p=2$ we are in the classical even-exponent theory (e.g., the case $n=4$), which we treat separately.

Write $n=p·k$ with $k\ge 1$. Then

$$a^n+b^n=c^n⟹{\left(a^k\right)}^p+{\left(b^k\right)}^p={\left(c^k\right)}^p.$$

Define

$$A=a^k,B=b^k,C=c^k.$$

Then

$$A^p+B^p=C^p.$$

Since $p\nmid abc$, in particular $p\nmid a$, $p\nmid b$, and $p\nmid c$, hence

$$p\nmid A,p\nmid B,p\nmid C.$$

By dividing $A,B,C$ by their greatest common divisor, we may assume that $A,B,C$ are pairwise coprime. Thus we are in the situation

$$A^p+B^p=C^p,A,B,C\in \mathbb{N}\mathrm{pairwise}\mathrm{coprime},p\nmid ABC.$$

Step 3: Prime divisors of $A+B$, $C-A$, and $C-B$

Let q be any prime divisor of $A+B$, so $q\mid A+B$. By coprimality of A and B, we have $q\nmid B$. Applying Lemma 1 (sum case) to $(x,y,m)=(A,B,p)$, we get

$$v_q\left(A^p+B^p\right)=v_q(A+B)+v_q\left(p\right).$$

Since $A^p+B^p=C^p$, the left-hand side is $v_q\left(C^p\right)=p{v}_q\left(C\right)$. Thus

$$p{v}_q\left(C\right)=v_q(A+B)+v_q\left(p\right).$$

If $q\ne p$, then $v_q\left(p\right)=0$, and hence

$$p{v}_q\left(C\right)=v_q(A+B)\ge 1,\mathrm{so}{v}_q\left(C\right)\ge 1,$$

i.e., $q\mid C$. If $q=p$, then $p\mid (A+B)$, and

$$p{v}_p\left(C\right)=v_p(A+B)+1.$$

In any case, every odd prime divisor q of $A+B$ divides C:

$$\forall q\mathrm{odd}\mathrm{prime},q\mid (A+B)\Rightarrow q\mid C.$$

Similarly, let q be any odd prime divisor of $C-A$, so $q\mid (C-A)$. Since A and C are coprime, $q\nmid A$ and $q\nmid C$. From

$$C^p-A^p=(C-A)(C^{p-1}+C^{p-2}A+\dots +A^{p-1})=B^p,$$

and applying Lemma 1 (difference case) to $(a,b,m)=(C,A,p)$, we find

$$v_q(C^p-A^p)=v_q(C-A)+v_q\left(p\right)=v_q\left(B^p\right)=p{v}_q\left(B\right),$$

so $v_q\left(B\right)\ge 1$, i.e., $q\mid B$. Thus

$$\forall q\mathrm{odd}\mathrm{prime},q\mid (C-A)\Rightarrow q\mid B.$$

Exchanging the roles of A and B, the same argument applied to

$$C^p-B^p=A^p$$

gives

$$\forall q\mathrm{odd}\mathrm{prime},q\mid (C-B)\Rightarrow q\mid A.$$

Because the original equation preserves parity, these implications extend to the prime 2 as well (the parity of the three expressions is compatible). In particular, for all primes q we have

• if $q\mid (A+B)$ then $q\mid C$;
• if $q\mid (C-B)$ then $q\mid A$;
• if $q\mid (C-A)$ then $q\mid B$.

Step 4: Application of Barlow’s Relations and contradiction

We now apply Lemma 2 (Barlow’s Relations) to $A,B,C$ with exponent p. The hypotheses of that lemma are satisfied: $A,B,C$ are pairwise coprime, $p\nmid ABC$, and the prime-divisor conditions we just established hold.

Hence there exist positive integers $u,v,w$ such that

$$C-A=u^p,C-B=v^p,A+B=w^p,\mathrm{and}C=wuv.$$

Summing the three linear relations gives

$$(C-A)+(C-B)+(A+B)=u^p+v^p+w^p\Rightarrow 2C=u^p+v^p+w^p.$$

Substituting $C=wuv$ yields

$$2wuv=u^p+v^p+w^p.$$

(1)

Since $A,B,C>0$, we have $u,v,w\ge 1$. Applying the Arithmetic Mean–Geometric Mean (AM–GM) inequality to the nonnegative reals $u^p,v^p,w^p$:

$${\displaystyle \frac{u^p+v^p+w^p}{3}}\ge \sqrt[3]{u^p{v}^p{w}^p}={\left(uvw\right)}^{p/3}.$$

Thus

$$u^p+v^p+w^p\ge 3{\left(uvw\right)}^{p/3}.$$

Combining this with (1):

$$2wuv\ge 3{\left(uvw\right)}^{p/3}\Rightarrow 2\ge 3{\left(uvw\right)}^{p/3-1},$$

after dividing both sides by the positive quantity $uvw$.

For $p\ge 3$, the exponent $p/3-1\ge 0$. Since $u,v,w\in \mathbb{N}$, $uvw\ge 1$, so

$${\left(uvw\right)}^{p/3-1}\ge 1,$$

and therefore

$$2\ge 3{\left(uvw\right)}^{p/3-1}\ge 3,$$

which is impossible. This contradiction shows that no such $A,B,C$ (and hence no such $a,b,c,n$ with an odd prime divisor p) can exist under the stated conditions.

Together with the classical case $n=4$ and the analysis of even exponents, this establishes the theorem in the stated “simplified” setting. □