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 establish an elementary proof of the nonexistence of solutions to the Fermat equation for prime exponents under a natural local restriction. Specifically, we assume that for an odd prime p, any hypothetical primitive solution triple $(a,b,c)$ to the equation

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

satisfies the inequality

$$a+b-c<2p.$$

The proof is remarkably direct: we show that the difference $\delta =a+b-c$ must satisfy $\delta \ge 2p$, contradicting the hypothesis. The argument combines three elementary facts. First, the binomial theorem ensures ${(a+b)}^p>a^p+b^p$, so $\delta >0$. Second, applying Fermat’s Little Theorem modulo p shows $p\mid \delta$, hence $\delta \ge p$. Third, parity considerations reveal that exactly one of $a,b,c$ is even, forcing $2\mid \delta$. Since $\delta$ is divisible by both 2 and p (with p odd), we obtain $\delta \ge 2p$, the desired contradiction. This fully elementary argument requires only basic congruence techniques and remains within the classical arithmetic framework.

2. Main Result

As usual, we write $d\mid n$ to mean that the integer d divides the integer n. We denote by $a\equiv b(modn)$ the congruence of a and b modulo n (i.e., $n\mid (a-b)$).

This is the main theorem.

Theorem 1 

(Fermat-type nonexistence under a local constraint). Let p be an odd prime. Suppose there exist positive integers $a,b,c$ such that

$$a^p+b^p=c^p,a,b,c\in \mathbb{N}\mathit{pairwise}\mathit{coprime}\mathit{and}a+b-c<2p.$$

Then no such integers $a,b,c$ exist.

Proof. 

Assume, for contradiction, that there exist positive integers $a,b,c$ and an odd prime p such that

$$a^p+b^p=c^p\mathrm{and}a+b-c<2p,$$

where $a,b,c$ are pairwise coprime.

Step 1: Establishing $\delta >0$ via the binomial theorem.

By the binomial theorem, expanding ${(a+b)}^p$ yields

$${(a+b)}^p=\sum _{k=0}^p\left({\displaystyle \genfrac{}{}{0pt}{}{p}{k}}\right)a^k{b}^{p-k}=a^p+b^p+\sum _{k=1}^{p-1}\left({\displaystyle \genfrac{}{}{0pt}{}{p}{k}}\right)a^k{b}^{p-k}.$$

Since all terms in the sum ${\sum }_{k=1}^{p-1}\left({\displaystyle \genfrac{}{}{0pt}{}{p}{k}}\right)a^k{b}^{p-k}$ are strictly positive (as $a,b>0$), we have

$${(a+b)}^p>a^p+b^p.$$

But $a^p+b^p=c^p$ by hypothesis, so

$${(a+b)}^p>c^p.$$

Since the function $x\mapsto x^p$ is strictly increasing on positive reals, this inequality implies $a+b>c$. Defining $\delta =a+b-c$, we conclude $\delta >0$.

Step 2: Showing $\delta \ge p$ via Fermat’s Little Theorem.

We now examine the equation $a^p+b^p=c^p$ modulo p. By Fermat’s Little Theorem, for any integer x, we have $x^p\equiv x(modp)$. Therefore,

$$a^p\equiv a(modp),b^p\equiv b(modp),c^p\equiv c(modp).$$

Substituting into $a^p+b^p=c^p$:

$$a+b\equiv c(modp)\Rightarrow a+b-c\equiv 0(modp)\Rightarrow p\mid \delta .$$

Since $\delta >0$ and $p\mid \delta$, we have $\delta \ge p$.

Step 3: Deducing $2\mid \delta$ from parity considerations.

Since p is an odd prime ($p\ge 3$), the exponent p is odd. For any integer x, the congruence $x^p\equiv x(mod2)$ holds because $x^p-x=x(x^{p-1}-1)$, which is even whether x is even (making the whole product even) or odd (making $x^{p-1}-1$ even). Reducing $a^p+b^p=c^p$ modulo 2:

$$a+b\equiv c(mod2).$$

If $a,b,c$ were all odd, we would have $1+1\equiv 1(mod2)$, i.e., $0\equiv 1(mod2)$, a contradiction. Thus at least one of $a,b,c$ is even. Since $a,b,c$ are pairwise coprime, at most one can be even (otherwise two would share the common factor 2). Therefore exactly one of $a,b,c$ is even and the other two are odd. In all cases,

$$a+b-c\equiv 0(mod2),$$

yielding $2\mid \delta$.

Step 4: Deriving the contradiction $\delta \ge 2p$.

We have established:

• $p\mid \delta$ (from Step 3), so $\delta \ge p$;
• $2\mid \delta$ (from Step 4);
• p is an odd prime, so $gcd(2,p)=1$.

Since $gcd(2,p)=1$ and both 2 and p divide $\delta$, their least common multiple $\mathrm{lcm}(2,p)=2p$ also divides $\delta$. Thus $\delta \ge 2p$. But this contradicts our hypothesis that $a+b-c=\delta <2p$.

Consequently, no such $a,b,c$ exist under the stated conditions. This completes the proof of the theorem. □