Suppose that for the positive integer $n$, $2^{n}+1$ is prime. What remainder can this prime give when divided by $240$?
Solution. We know that $n$ is a positive integer, and $2^{n}+1$ is a prime number. If there existed an odd prime number $p$ such that $p \mid n$, then there would be an $x \in \mathbb{N}$ such that $n=p \cdot x$.

Thus, $2^{n}+1=2^{p \cdot x}+1=\left(2^{x}\right)^{p}+1^{p}$. According to the known identity,

$$
2^{x}+1 \mid\left(2^{x}\right)^{p}+1^{p}
$$

and $1<2^{x}+1<2^{n}+1$, so it would be a proper divisor, meaning $2^{n}+1$ could not be a prime number. We have reached a contradiction, so $n$ cannot have an odd divisor, hence $n=2^{k}$.

Let's examine the remainders when dividing by 240 for exponents of this form:

$$
\begin{aligned}
& 2^{2^{0}}+1=3 \equiv 3 \quad(\bmod 240) \\
& 2^{2^{1}}+1=5 \equiv 5 \quad(\bmod 240) \\
& 2^{2^{2}}+1=17 \equiv 17 \quad(\bmod 240) \\
& 2^{2^{3}}+1=257 \equiv 17 \quad(\bmod 240) \\
& 2^{2^{4}}+1=65537 \equiv 17 \quad(\bmod 240)
\end{aligned}
$$

Conjecture: the remainder will always be 17 from this point onward.

$$
240=2^{4} \cdot 3 \cdot 5
$$

So we need to examine what remainders the numbers of the form $2^{2^{k}}+1$ give when divided by $2^{4}$, 3, and 5.

Let $A=2^{2^{k}}+1$.

If $k \geq 2$, then $2^{4} \mid 2^{2^{k}}$, so $2^{4}\left|A-1 \Rightarrow 2^{4}\right| A-1-16 \Rightarrow 2^{4} \mid A-17$.

$$
\begin{aligned}
A= & 2^{2^{k}}+1=(3-1)^{2^{k}}+1 \equiv(-1)^{2^{k}}+1 \equiv 2 \quad(\bmod 3) \Longrightarrow \\
& \Longrightarrow 3|A-2 \Longrightarrow 3| A-2-15 \Longrightarrow 3 \mid A-17
\end{aligned}
$$

The last digits of powers of 2 for positive exponents are $2,4,8,6,2,4,8,6$ and so on. Therefore, the last digit of $A=2^{2^{k}}+1$ is 7, so the last digit of $A-17$ is 0, meaning $5 \mid A-17$.

Thus, for $k \geq 2$, the prime number $2^{n}+1$ will always have a remainder of 17 when divided by 240.

Therefore, the possible remainders are 3, 5, and 17.