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 the Diophantine equation

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

has no solutions in positive integers 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 [2], Sophie Germain [3], and Kummer [4] resolved important special cases, yet a complete proof remained elusive.

The modern breakthrough came in 1994, when Andrew Wiles established the full theorem using deep results from the theory of elliptic curves and modular forms [5]. His work, later strengthened by Ribet and others [6], revolutionized modern number theory and relied on sophisticated machinery far removed from the classical arithmetic methods available in Fermat’s time.

Despite this achievement, the search for an elementary proof—one relying only on classical number-theoretic tools—has persisted. Such a proof would illuminate the inherent arithmetic structure of the Fermat equation and provide insight into why the equation admits no nontrivial solutions.

In this article we present an elementary proof of Fermat’s Last Theorem for all exponents $n\ge 3$. The argument reduces the general case to an odd prime exponent and then applies a structural lemma—Barlow’s Relations—which describes the precise algebraic form that any hypothetical solution must satisfy. When combined with p-adic valuation techniques, these relations impose divisibility and size constraints that are mutually incompatible. The resulting contradiction eliminates all possible solutions, thereby establishing the theorem without recourse to modern analytic or geometric methods.

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

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

Then the integers $a,b,c$ must satisfy the following structural conditions:

• Case 1 ($p\nmid c$):  There exist positive integers $u,v,w$ such that

  $$c-a=u^p,c-b=v^p,a+b=w^p.$$
• Case 2 ($p\mid c$):  There exist positive integers $u,v,w$ and an integer $k\ge 1$ such that

  $$c-a=u^p,c-b=v^p,a+b=p^{kp-1}{w}^p,$$
  with $gcd(w,p)=1$.

Furthermore, if for every prime q we have

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

and with the structural relations

$$c-a=u^p,c-b=v^p,a+b=\left\{\begin{array}{cc}{w}^p\hfill & \mathit{if}p\nmid c,\hfill \\ p^{kp-1}{w}^p\hfill & \mathit{if}p\mid c,\hfill \end{array}\right.$$

we prove the inequality

$$1+{\displaystyle \frac{2c-3}{p}}<uvw.$$

Proof. 

We use your detailed structural analysis and split into the two cases according to whether p divides c or not.

1. 

Factorization of $a^p+b^p$

Write

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

where

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

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

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

Let

$$d=gcd(a+b,Q(a,b\left)\right).$$

Then $d\mid p{a}^{p-1}$. Since $gcd(a,b)=1$, we have $gcd(a+b,a)=1$, hence $gcd(a+b,a^{p-1})=1$. Thus

$$d\mid p.$$

Now we split into the two cases.

Case 1: 

$p\nmid c$

From $a^p+b^p=c^p$ and $p\nmid c$, we have $p\nmid (a^p+b^p)$. If $p\mid (a+b)$, then from the congruence above we get $p\mid Q(a,b)$, hence

$$p^2\mid (a+b)Q(a,b)=c^p,$$

so $p\mid c$, contradiction. Therefore

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

Since

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

is a perfect p-th power and the two factors are coprime, the valuation argument applies:

For any prime q,

$$v_q(a+b)+v_q\left(Q(a,b)\right)=v_q\left(c^p\right)=p{v}_q\left(c\right).$$

Because $gcd(a+b,Q)=1$, at most one of $v_q(a+b),v_q\left(Q\right)$ is nonzero. Thus each nonzero valuation is a multiple of p. Hence there exist integers $w,z_1\ge 1$ such that

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

and then

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

So in Case 1 we already have

$$a+b=w^p.$$

We now show that $c-a$ and $c-b$ are also perfect p-th powers.

• Factorizations of $c^p-a^p$ and $c^p-b^p$

Write

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

where

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

Modulo $c-a$, we have $c\equiv a$, hence

$$S(c,a)\equiv p{a}^{p-1}(modc-a).$$

Let

$$d_1=gcd(c-a,S(c,a)).$$

Then $d_1\mid p{a}^{p-1}$. Since $gcd(a,c)=1$, we have $gcd(c-a,a)=1$, hence $gcd(c-a,a^{p-1})=1$. Thus

$$d_1\mid p.$$

From $c^p-a^p=b^p$ and $gcd(a,b)=gcd(b,c)=1$, we have $p\nmid b$ (otherwise $p\mid a^p+b^p=c^p$ would force $p\mid a,b,c$, contradicting coprimality). Hence $p\nmid (c^p-a^p)$. If $p\mid (c-a)$, then by LTE (difference case),

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

so $p^2\mid b^p$, hence $p\mid b$, contradiction. Thus

$$p\nmid (c-a),d_1=1.$$

Since

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

is a perfect p-th power and the factors are coprime, the same valuation argument yields

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

A parallel argument applied to

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

gives

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

Thus in Case 1 we have shown:

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

as required.

Case 2: 

Case 2: $p\mid c$

Now suppose $p\mid c$. Since $gcd(a,c)=gcd(b,c)=1$, we must have $p\nmid a$ and $p\nmid b$. From

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

and $p\mid c$, we have $p\mid a^p+b^p$. Applying LTE (sum case) with exponent p gives

$$v_p(a^p+b^p)=v_p(a+b)+v_p\left(p\right)=v_p(a+b)+1.$$

But also

$$v_p(a^p+b^p)=v_p\left(c^p\right)=p{v}_p\left(c\right).$$

Let $v_p\left(c\right)=k\ge 1$. Then

$$pk=v_p(a+b)+1⟹v_p(a+b)=pk-1.$$

Write

$$a+b=p^{pk-1}M,$$

with $p\nmid M$. From

$$(a+b)Q(a,b)=c^p=p^{pk}{C}_0^p$$

for some integer $C_0$ with $p\nmid C_0$, we get

$$p^{pk-1}M·Q(a,b)=p^{pk}{C}_0^p⟹MQ(a,b)=p{C}_0^p.$$

Since $p\nmid M$, we must have $v_p\left(Q(a,b)\right)=1$, and then

$$Q(a,b)=pN,MN=C_0^p,$$

with $p\nmid N$. Because $gcd(M,N)=1$ and $MN$ is a perfect p-th power, the valuation argument shows that both M and N are perfect p-th powers. Thus there exist integers $w,z_1\ge 1$ such that

$$M=w^p,N=z_1^p,$$

with $gcd(w,p)=1$. Therefore

$$a+b=p^{pk-1}{w}^p,$$

as claimed in Case 2.

The analysis of $c-a$ and $c-b$ proceeds exactly as in Case 1. The previous computations did not use the assumption $p\nmid c$; they only used:

• $gcd(a,c)=1$ to get $gcd(c-a,a)=1$,
• and $p\nmid b$ to rule out $p\mid (c-a)$ via LTE.

In Case 2 we still have $gcd(a,c)=1$ and $p\nmid b$ (since $p\mid c$ and $gcd(b,c)=1$), so the same argument applies and yields

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

for some positive integers $u,v$.

Thus in Case 2 we have shown:

$$c-a=u^p,c-b=v^p,a+b=p^{kp-1}{w}^p,gcd(w,p)=1,$$

with $k=v_p\left(c\right)\ge 1$.

• A Key Inequality

Recall that from the structural relations we have $w\mid c$, $u\mid b$, $v\mid a$; write $c=w\tilde{c}$ with $\tilde{c}\in \mathbb{N}$. Pairwise coprimality of $(a,b,c)$ implies that $u,v,w$ are pairwise coprime. Moreover, $w\ge 2$ because $a+b=w^p$ (or $a+b=p^{kp-1}{w}^p$) and $a,b\ge 1$.

From the three structural relations we obtain the linear identity

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

(1)

We first establish the auxiliary inequality

$$uvw\ge u+v+w-2.$$

(2)

If $u=v=1$, then $w\ge 2$ and $uvw=w=1+1+w-2=u+v+w-2$, so (2) holds with equality. But the case $u=v=1$ is impossible under our hypotheses: it would imply $a=c-1$ and $b=c-1$, hence $a=b$, contradicting $gcd(a,b)=1$. Thus in any admissible triple we have either $u\ge 2$ or $v\ge 2$ (or both). If both $u\ge 2$ and $v\ge 2$, then $uvw\ge 4w$ and $u+v+w-2\le 2w+w-2=3w-2$; since $4w\ge 3w-2$ for $w\ge 2$, inequality (2) holds. If one of $u,v$ equals 1 and the other is at least 2, say $u=1$ and $v\ge 2$, then $uvw=vw$. From the identity $2v{z}_3=w^p+v^p-1$ (obtained during the proof of the structural relations) we deduce $w^p\ge v^p+1$, hence $w>v$ and consequently $w\ge v+1$. Therefore

$$uvw=vw\ge v(v+1)>v+1+v-2=v+w-1=u+v+w-2,$$

so (2) holds strictly. Thus in all admissible cases we have $uvw\ge u+v+w-2$, with equality only when $u=v=1$, which is excluded.

Next we use the elementary inequality $x^p\ge px-(p-1)$ for integers $x\ge 1$, which follows from Bernoulli’s inequality (or by induction). Applying it to $u,v,w$ gives

$$u^p+v^p+w^p\ge p(u+v+w)-3(p-1).$$

(3)

Combining (1), (2) and (3) we obtain

$$2c=u^p+v^p+w^p\ge p(u+v+w)-3(p-1)\ge p(uvw+2)-3(p-1)=puvw+2p-3p+3=puvw-p+3.$$

Hence

$$puvw\le 2c+p-3⟹uvw\ge {\displaystyle \frac{2c}{p}}+1-{\displaystyle \frac{3}{p}}.$$

(4)

Since $p\ge 3$, we have $1-{\displaystyle \frac{3}{p}}\ge 0$, and (4) can be rewritten as

$$uvw\ge 1+{\displaystyle \frac{2c-3}{p}}.$$

The equality case in (4) would require equality in both (2) and (3). Equality in (2) forces $u=v=1$, which is impossible as noted. Therefore the inequality is strict:

$$uvw>1+{\displaystyle \frac{2c-3}{p}}.$$

This completes the proof of the key inequality. □