Let $a, b$ be relatively prime integers and let $p$ be a prime number.
Suppose that we have
\[a^{2^n}+b^{2^n}\equiv 0 \pmod{p}\]
for some positive integer $n$.

Since $a$ and $b$ are relatively prime, at least one of them is relatively prime to $p$.
Without loss of generality let us assume that $b$ and $p$ are relatively prime.

Then the given equality becomes
\begin{align*}
a^{2^n}\equiv -b^{2^n} \pmod{p} \\
\iff \left( \frac{a}{b}\right)^{2^n} \equiv -1 \pmod{p}.
\end{align*}
Taking square of both sides we obtain
\[\left( \frac{a}{b}\right)^{2^{n+1}} \equiv 1 \pmod{p}.\]

Now, we can think of these congruences as equalities of elements in the multiplicative group $(\Z/p\Z)^{\times}$ of order $p-1$:
\[ \left( \frac{a}{b}\right)^{2^n} = -1 \text{ and } \left( \frac{a}{b}\right)^{2^{n+1}} =1 \text{ in } (\Z/p\Z)^{\times}.\]

Note that the second equality yields that the order of the element $a/b$ divides $2^{n+1}$.
On the other hand, the first equality implies that any smaller power of $2$ is not the order of $a/b$.
Thus, the order of the element $a/b$ is exactly $2^{n+1}$.

In general, the order of each element divides the order of the group.
(This is a consequence of Lagrange’s theorem.)

Since the order of the group $(\Z/p\Z)^{\times}$ is $p-1$, it follows that $2^{n+1}$ divides $p-1$.
This completes the proof.

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