Using the factorization of the difference of squares, $x^{2}-y^{2}=(x-y)(x+y)$, we can write

$$
x^{2^{n}}-y^{2^{n}}=\left(x^{2^{n-1}}+y^{2^{n-1}}\right)\left(x^{2^{n-1}}-y^{2^{n-1}}\right) \text {. }
$$

a) Explain why $3^{2^{2018}}-2^{2^{2018}}$ can be written as a product of 2018 distinct integers greater than 1.

b) Verify that $3^{2^{n}}-1=\left(3^{2^{n-1}}+1\right)\left(3^{2^{n-2}}+1\right)\left(3^{2^{n-3}}+1\right) \ldots\left(3^{2}+1\right)\left(3^{1}+1\right)\left(3^{1}-1\right)$.

c) Using the previous item, verify that $2^{n+1}$ is a divisor of $3^{2^{n}}-1$.

d) Knowing the factorization

$$
x^{m}-y^{m}=(x-y)\left(x^{m-1}+x^{m-2} y+x^{m-3} y^{2}+\ldots+x y^{m-2}+y^{m-2}\right)
$$

find a positive integer $n$ with more than 2018 positive divisors such that $3^{n-1}-2^{n-1}$ is a multiple of $n$.
Solution

a) Using the factorization of the difference of squares, we have:

$$
\begin{aligned}
x^{2^{n}}-y^{2^{n}} & =\left(x^{2^{n-1}}+y^{2^{n-1}}\right)\left(x^{2^{n-1}}-y^{2^{n-1}}\right) \\
x^{2^{n-1}}-y^{2^{n-1}} & =\left(x^{2^{n-2}}+y^{2^{n-2}}\right)\left(x^{2^{n-2}}-y^{2^{n-2}}\right) \\
x^{2^{n-2}}-y^{2^{n-2}} & =\left(x^{2^{n-3}}+y^{2^{n-3}}\right)\left(x^{2^{n-3}}-y^{2^{n-3}}\right) \\
& \cdots \\
x^{2}-y^{2} & =(x+y)(x-y)
\end{aligned}
$$

Multiplying all the equations and canceling the repeated factors, we obtain

$$
x^{2^{n}}-y^{2^{n}}=\left(x^{2^{n-1}}+y^{2^{n-1}}\right)\left(x^{2^{n-2}}+y^{2^{n-2}}\right) \ldots\left(x^{2}+y^{2}\right)(x+y)(x-y)
$$

Thus, in general, substituting \( x=3 \) and \( y=2 \), \( 3^{2^{n}}-2^{2^{n}} \) is the product of \( n \) factors of the form \( 2^{2^{i}}+3^{2^{i}} \) with \( i \) belonging to the set \( \{0,1,2, \ldots, n-1\} \).

b) In the factorization obtained in the previous item, it suffices to set \( x=3 \) and \( y=1 \).

c) Since each term of the form \( 3^{2^{i}}+1 \) has the factor 2 because it is even, we can conclude that the product of all numbers of the form \( 3^{2^{i}}+1 \), with \( i \in \{1,2, \ldots, n-1\} \), is a multiple of \( 2^{n-1} \). Moreover, the factor \( 3^{2^{0}}+1=2^{2} \) has two factors of 2. Thus, using the expression found in the previous item, we can conclude that \( 2^{n-1} \cdot 2^{2}=2^{n+1} \) is a divisor of \( 3^{2^{n}}+1 \).

d) Let \( n=3^{2^{2017}}-2^{2^{2017}} \). From the first item, all 2019 numbers in the list

$$
1,2^{2^{0}}+3^{2^{0}}, 2^{2^{1}}+3^{2^{1}}, \ldots, 3^{2^{2017}}+2^{2^{2017}}
$$

are divisors of \( n \). From the given factorization, we can conclude that if \( x \) and \( y \) are integers, then \( x-y \) is a divisor of \( x^{m}-y^{m} \). Setting \( x=2^{2^{2017}} \) and \( y=3^{2^{2017}} \), we obtain that \( n=3^{2^{2017}}-2^{2^{2017}} \) is a divisor of \( 3^{2^{2017} m}-2^{2^{2017} m} \), for any positive integer \( m \). Note now that

$$
n-1=\left(3^{2^{2017}}-1\right)-2^{2^{2017}}
$$

From the third item, \( 3^{2^{2017}}-1 \) is a multiple of \( 2^{2017} \). Since \( 2017 < 2^{2017} \), we also have that \( 2^{2017} \) is a divisor of \( 2^{2^{2017}} \). Thus, \( \left(3^{2^{2017}}-1\right)-2^{2^{2017}} \) is a number of the form \( 2^{2017} m \). Therefore, \( n \) divides \( 3^{n-1}-2^{n-1} \) and, moreover, has at least 2019 positive divisors.