Exercise 1.2.51* (Fermat's prequel)

Question

Show that if $(a,b)=1$ and $p$ is an odd prime, then
$$
\left(a+b, \frac{a^p+b^p}{a+b} ight) = 1 	ext{ or } p.
$$

Richard Ganaye · Accepted Answer

Note that this is a generalization of Exercise 1.2.50. We use the same method (cut and paste).
\begin{proof} Write $d = a+b \wedge \frac{a^p+b^p}{a+b} $.

Then $d \mid a + b$ and 
$$d \mid  \frac{a^p+b^p}{a+b}  = \sum_{k=0}^{p-1} (-1)^k a^{p-1-k}b^{k} = a^{p-1} - a^{p-2}b + \cdots + (-1)^k a^{p-1-k}b^{k}  + \cdots + b^{p-1}.$$
Therefore $-b \equiv a \pmod d$, thus
\begin{align*}
0 & \equiv \sum_{k=0}^{p-1} a^{p-1-k} (-b)^k\
&\equiv \sum_{k=0}^{p-1} a^{p-1-k} a^k = pa^{p-1} \pmod d,\
\end{align*}
so
$$
\left\{
\begin{array}{ll}
d \mid a + b,\
d \mid pa^{p-1}.
\end{array}
ight.
$$
By hypothesis, $a\wedge b = 1$, thus $a \wedge a+b = 1$, and $a^{p-1} \wedge a+b = 1$.

Since $d \mid a+b$, then $d \wedge a^{p-1} = 1$ (for every integer $c$, if $c \mid d$ and $c \mid a^{p-1}$, then $c \mid a+b$, so $c \mid a^{p-1} \wedge a+b = 1$).

From $d \mid pa^{p-1}$ and $d \wedge a^{p-1} = 1$, we deduce $d \mid p$, where $d\geq 0$.

This shows that $d = 1$ or $d =p$.

\end{proof}

Note: This is a lemma for the first case of the (big) Fermat's Theorem. If $c^p =a ^p + b^p$, where $p 
mid abc$
then $c^p = (a+b)\frac{a^p+b^p}{a+b} $. Since $p 
mid c$, the preceding result shows that $(a+b) \wedge \frac{a^p+b^p}{a+b}  = 1$, therefore both factors are $p$-th powers. This is the beginning of a great story.

Exercise 1.2.51* (Fermat's prequel)

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Exercise 1.2.51* (Fermat's prequel)

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