Putnam 2018 - B3

Problem: Find all positive integers $n<10^{100}$ for which simultaneously $n$ divides $2^n$, $n-1$ divides $2^n-1$ and $n-2$ divides $2^n-2$.
Solution: Since $n|2^{n}\implies n=2^{k}$ for some $k\ge 0$. Substituting into the second divisibility, $2^{k}-1|2^{2^{k}}-1$. Now we use the result $gcd(2^{a}-1,2^{b}-1)=2^{gcd(a,b)}-1$ to see $gcd(2^{2^{k}}-1,2^{k}-1)=2^{gcd(2^{k},k)}-1=2^{k}-1$, therefore $gcd(2^{k},k)=k$ and $k|2^{k}$, so $k=2^{m}$. Therefore, $n=2^{2^{m}}$. Substituting into the final divisibility,
$2^{2^{m}}-2|2^{2^{2^{m}}}-2\implies 2^{2^{m}-1}-1|2^{2^{2^{m}-1}}-1$. Using the exponential divisible sequence property twice, $gcd(2^{2^{2^{m}-1}}-1,2^{2^{m}-1}-1)=2^{gcd(2^{2^{m}-1},2^{m}-1)}-1=2^{2^{gcd(2^{m},m)}-1}-1=2^{2^{m}-1}-1,$
therefore $gcd(2^{m},m)=m$ and $m=2^{l}$ for some $l\ge 0$. Therefore $n=2^{2^{2^{l}}}$ for $l=0,1,2,3$ giving the solutions $n=4,16,65536,2^{256}$. Note that $2^{256}<2^{300}=8^{100}<10^{100}$, therefore these all satisfy $n<10^{100}$. Clearly for $l\ge 4,n=2^{2^{2^{l}}}\ge 2^{2^{16}}=2^{65536}>>10^{100}$.