Question 94
Using the square difference factorization, $x^{2}-y^{2}=(x-y)(x+y)$, 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 an integer 2018 product. greater than 1 and distinct.
b) Check 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, check 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 $n$.
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{.}
$$
a) Explain why $3^{2^{2018}}-2^{2^{2018}}$ can be written as an integer 2018 product. greater than 1 and distinct.
b) Check that $3^{2^{n}}-1=\left(3^{2^{n-1}}+1\right)\left(3^{2^{n-2}}+1\right)\left(3^{2^{n-3}}+1\right) \ldots\left(3^{2}+1\right)\left(3^{1}+1\right)\left(3^{1}-1\right)$.
c) Using the previous item, check that $2^{n+1}$ is a divisor of $3^{2^{n}}-1$.
d) Knowing the factorization $$
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 $n$.