1. Introduction

2. Background and Ancillary Results

3. Main Result

• the consideration of an odd prime p as the selected exponent;
• the coprimality of $a,b,c$;
• and the condition $a,b,c>1$ on account of Catalan’s conjecture, proven by Mihăilescu in [10].

Case 1:

Suppose that $a,b,c$ are pairwise coprime with p. Using the Proposition 2.3 we notice that

$$p\mid c^p-b^p-(c-b)\Rightarrow p\mid a^p-(c-b).$$

(2)

First we start with an equivalent expression of (1)

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

Substituting $x=c$, $y=b$ and using that p is odd,

$$c^p-b^p=(c-b)·\sum _{k=0}^{p-1}{c}^k·b^{p-1-k}=a^p$$

(3)

by Proposition 2.4. That is equivalent to

$$\begin{array}{cc}\hfill (c-b)·\sum _{k=0}^{p-1}{c}^k·b^{p-1-k}& =(c-b)·\left(1-1+\sum _{k=0}^{p-1}{c}^k·b^{p-1-k}\right)\hfill \\ \hfill & =(c-b)+(c-b)·\left(-1+\sum _{k=0}^{p-1}{c}^k·b^{p-1-k}\right).\hfill \end{array}$$

So, we would have

$$(c-b)·\left(-1+\sum _{k=0}^{p-1}{c}^k·b^{p-1-k}\right)=a^p-(c-b)$$

from (3). If the prime number p divides $(c-b)$, then $p\mid a^p$ and thus, a is divisible by p. If p does not divide a, then this implies

$$p\mid \left(-1+\sum _{k=0}^{p-1}{c}^k·b^{p-1-k}\right)$$

according to Proposition 2.2 and properties of (2). However, we can see that

$$\begin{array}{cc}\hfill \left(-1+\sum _{k=0}^{p-1}{c}^k·b^{p-1-k}\right)& =\left(c^{p-1}-1+b·\sum _{k=0}^{p-2}{c}^k·b^{p-2-k}\right)\hfill \\ \hfill & =\left(b^{p-1}-1+c·\sum _{k=0}^{p-2}{c}^k·b^{p-2-k}\right).\hfill \end{array}$$

We know that

$$p\mid c^{p-1}-1,p\mid b^{p-1}-1$$

by Proposition 2.3 since p and $c,b$ are pairwise coprime. Consequently, we obtain that ($p\mid {\sum }_{k=0}^{p-2}{c}^k·b^{p-2-k}$) or ($p\mid c$ or $p\mid b$) by Proposition 2.2. It is not possible that ($p\mid c$ or $p\mid b$) whenever p and $c,b$ are pairwise coprime and therefore, it would be necessary that ($p\mid {\sum }_{k=0}^{p-2}{c}^k·b^{p-2-k}$). In virtue of (3), we would have

$$\begin{array}{cc}\hfill (c-b)·\sum _{k=0}^{p-1}{c}^k·b^{p-1-k}& =(c-b)·\left(c^{p-1}+b^{p-1}+\sum _{k=0}^{p-2}{c}^k·b^{p-2-k}\right)\hfill \\ \hfill & =(c-b)·\left(2+\left(c^{p-1}-1+b^{p-1}-1\right)+\sum _{k=0}^{p-2}{c}^k·b^{p-2-k}\right)\hfill \end{array}$$

which is

$$(c-b)+(c-b)·\left(\left(c^{p-1}-1+b^{p-1}-1\right)+\sum _{k=0}^{p-2}{c}^k·b^{p-2-k}\right)=a^p-(c-b).$$

By Proposition 2.3 and (2), we can further deduce that a is divisible by p because p would divide $(c-b)$. Since $a,b,c$ are pairwise coprime with p, we reach a contradiction.

Case 2:

Suppose that $b,c$ are pairwise coprime with p and a is divisible by p. By Proposition 2.3, we can see that

$$p\mid a^p+b^p-(a+b)\Rightarrow p\mid c^p-(a+b).$$

(4)

Substituting $x=a$, $y=-b$ and using that p is odd,

$$a^p+b^p=(a+b)·\sum _{k=0}^{p-1}{a}^k·{(-b)}^{p-1-k}=c^p$$

(5)

by Proposition 2.4. That would be

$$\begin{array}{cc}\hfill (a+b)·\sum _{k=0}^{p-1}{a}^k·{(-b)}^{p-1-k}& =(a+b)·\left(1-1+\sum _{k=0}^{p-1}{a}^k·{(-b)}^{p-1-k}\right)\hfill \\ \hfill & =(a+b)+(a+b)·\left(-1+\sum _{k=0}^{p-1}{a}^k·{(-b)}^{p-1-k}\right).\hfill \end{array}$$

After that, we check

$$(a+b)·\left(-1+\sum _{k=0}^{p-1}{a}^k·{(-b)}^{p-1-k}\right)=c^p-(a+b)$$

from (5). If the prime number p divides $(a+b)$, then $p\mid c^p$ and thus, c is divisible by p. If p does not divide c, then this implies

$$p\mid \left(-1+\sum _{k=0}^{p-1}{a}^k·{(-b)}^{p-1-k}\right)$$

according to Proposition 2.2 and properties of (4). Nevertheless, we can see that

$$\left(-1+\sum _{k=0}^{p-1}{a}^k·{(-b)}^{p-1-k}\right)=\left(b^{p-1}-1+a^{p-1}+\sum _{k=0}^{p-2}{a}^k·{(-b)}^{p-2-k}\right).$$

We know that

$$p\mid a^{p-1},p\mid b^{p-1}-1$$

by Proposition 2.3 since $p\mid a$ and p and b are pairwise coprime. Consequently, we obtain that

$$p\mid \sum _{k=0}^{p-2}{a}^k·{(-b)}^{p-2-k}.$$

Hence, it is enough to show that

$$\sum _{k=0}^{p-2}{a}^k·{(-b)}^{p-2-k}=-b^{p-2}+a·m$$

for $m\in \mathbb{Z}$. Since $p\mid a·m$, then we can further deduce that b is divisible by p due to $p\mid -b^{p-2}$. Since $b,c$ are pairwise coprime with p, we reach a contradiction.

Case 3:

Suppose that $a,c$ are pairwise coprime with p and b is divisible by p. Following the same steps as the above case mutatis mutandis, and exploiting the symmetry of the left-hand side of (1) with respect to a and b, we get another contradiction after of applying the same arguments to the value of b instead of choosing the number a as a multiple of p.

Case 4:

Suppose that $a,b$ are pairwise coprime with p and c is divisible by p. Same as Case 3.

Case 5:

Finally, we arrive at the following conclusion: Natural numbers $a,b,c$ share p as a common prime factor. However, this poses a contradiction with the pairwise coprimality of $a,b,c\in \mathbb{N}$ assumed from the outset in (1).

4. Conclusion

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