Prove that for every positive integer $n$, there is a positive integer $m$ so that $2^n | (19^m-97)$.🔑
Proof: We prove it by induction.
Base Case: For $n=1,2,3$, we have $$n=1, m=1 \Longrightarrow 2^1|(19^1-97)$$ $$n=2, m=2 \Longrightarrow 2^2|(19^2-97)$$ $$n=3,m=2 \Longrightarrow 2^3|(19^3-97)$$
Inductive Step: Assume for $n$, where $n>3$, there is a positive integer $m$ so that $2^n | (19^m-97)$. This means $$19^m-97=x\cdot 2^{n}$$ where $x$ is an integer.
Case 1: If $x$ is even, then $2^{n+1} | (x\cdot 2^n)$, i.e. $2^{n+1}|(19^m-97)$, we are done.
Case 2: If $x$ is odd. Because for integer $n>2$, we have $$19^{2^{n-2}}=1+y\cdot 2^n$$ where $y$ is an odd integer (see the proof here). Therefore, $$19^{m+2^{n-2}}-97=19^{m}\cdot 19^{2^{n-2}}-97=(97+x\cdot 2^n)(1+y\cdot 2^n)-97$$ $$\ \ \ \ \ \ \ \ =97+97\cdot y\cdot 2^n+x\cdot 2^n+x\cdot y\cdot 2^{2n}-97$$ $$=(x+97\cdot y)\cdot 2^n+x\cdot y\cdot 2^{2n}\ \ \ \ \ \ \ \ \ \ \ \ $$
Because $x$ and $y$ are odd integers, $x+97\cdot y$ is even, thus $2^{n+1}|(x+97\cdot y)\cdot 2^n$. Additionally, because $n>3$, $2n>n+1$, thus $2^{n+1}|x\cdot y\cdot 2^{2n}$. Therefore, we have $$2^{n+1}|\Big{(}(x+97\cdot y)\cdot 2^n+x\cdot y\cdot 2^{2n}\Big{)}$$ i.e. $$2^{n+1}|(19^{m+2^{n-2}}-97)$$
Combining Case 1 and 2, we complete the inductive step.
By induction, the problem is proved.