$a_{n+1} =\frac{a_n}{2} (modp)$, then $a_n=\frac{1}{2^{n-1}} (modp)$. Now if $p \mid gcd(2^n+1, 2^{t}-1)$, where $n$ is the Least number which satisfies $2^x=-1(mod p)$, and $t=ord_p(2) \mid 2019$, $2^{2n}=1(modp)$, then $t \mid 2n$, $t \mid n$ because $t=1(mod2)$, then $2^n=1=-1(modp)$, a contradiction.

WypHxr wrote: $a_{n+1} =\frac{a_n}{2} (modp)$, then $a_n=\frac{1}{2^{n-1}} (modp)$. Now if $p \mid gcd(2^n+1, 2^{t}-1)$, where $n$ is the Least number which satisfies $2^x=-1(mod p)$, and $t=ord_p(2) \mid 2019$, $2^{2n}=1(modp)$, then $t \mid 2n$, $t \mid n$ because $t=1(mod2)$, then $2^n=1=-1(modp)$, a contradiction. I wonder how to come up with $a_{n+1} =\frac{a_n}{2} (modp)$.

It's equivalent to prove that $2a_{n+1}\equiv{a_n}\pmod{p}$ We prove this by induction on $n$. $n=0$ gives $2a_1=2=a_0\pmod{p}$ which is true. Assume that for $n$ we have $2a_{n+1}\equiv{a_n}\pmod{p}$, then : $2a_{n+2}\equiv{2a_{n+1}+\frac{p^2-1}{2}a_n} \equiv{a_{n+1}+\frac{2a_{n+1}+(p^2-1)a_n}{2}}\pmod{p}$ And $2a_{n+1}+(p^2-1)a_n\equiv{2a_{n+1}-a_n} \equiv {0}\pmod{p} $ and $p$ is odd so $2a_{n+2}\equiv {a_{n+1}} \pmod{p}$ which achieve the induction!!

Zsigmondy123 wrote: It's equivalent to prove that $2a_{n+1}\equiv{a_n}\pmod{p}$ We prove this by induction on $n$. $n=0$ gives $2a_1=2=a_0\pmod{p}$ which is true. Assume that for $n$ we have $2a_{n+1}\equiv{a_n}\pmod{p}$, then : $2a_{n+2}\equiv{2a_{n+1}+\frac{p^2-1}{2}a_n} \equiv{a_{n+1}+\frac{2a_{n+1}+(p^2-1)a_n}{2}}\pmod{p}$ And $2a_{n+1}+(p^2-1)a_n\equiv{2a_{n+1}-a_n} \equiv {0}\pmod{p} $ and $p$ is odd so $2a_{n+2}\equiv {a_{n+1}} \pmod{p}$ which achieve the induction!! But how to observe $2a_{n+1}\equiv{a_n}\pmod{p}$.

Use induction to show that: $a_{n}=(\frac{p+1}{2})^n+(\frac{1-p}{2})^n$ So we need to show that for any $n$ we can't have $p|2^n+2$ this is equivalent to $2^{n-1} \equiv -1 \pmod{p} $ which means $Ord_p(2)$ is even which contradicts $p|2^{2019}-1$.

WindyLi wrote: WypHxr wrote: $a_{n+1} =\frac{a_n}{2} (modp)$, then $a_n=\frac{1}{2^{n-1}} (modp)$. Now if $p \mid gcd(2^n+1, 2^{t}-1)$, where $n$ is the Least number which satisfies $2^x=-1(mod p)$, and $t=ord_p(2) \mid 2019$, $2^{2n}=1(modp)$, then $t \mid 2n$, $t \mid n$ because $t=1(mod2)$, then $2^n=1=-1(modp)$, a contradiction. I wonder how to come up with $a_{n+1} =\frac{a_n}{2} (modp)$. Basically you are given a linear recursion in the field \(\mathbb F_p\). In other words you can reduce the equation modulo \(p\) to get \[a_{n+1} = a_{n} - \frac{1}{4} a_{n-1}.\]Now you can use the generic characteristic polynomial method to guess a closed form for \(a_n\); the polynomial here is \((x-1/2)^2\), so you have a double root. Such recursions have a closed form of \((An + B)(1/2)^n\), and from \(a_0 = 2\) and \(a_1 = 1\) you get \(A=0\) and \(B=2\). This gives \(a_n \equiv 2^{1-n}\), and now you prove that this indeed works by induction on \(n\).

Taha1381 wrote: Use induction to show that: $a_{n}=(\frac{p+1}{2})^n+(\frac{1-p}{2})^n$ So we need to show that for any $n$ we can't have $p|2^n+2$ this is equivalent to $2^{n-1} \equiv -1 \pmod{p} $ which means $Ord_p(2)$ is even which contradicts $p|2^{2019}-1$. I think it is not true