Number Theory Challenge – 12/20/2025

Proof: We prove it by induction.

Base Case: When $n=1$, we have $$19^{2^n}=361=1+45\cdot 2^3=1+45\cdot 2^{n+2}$$ Therefore we prove the case when $n=1$, is true with $m=45$ as an odd integer.

Inductive Step: Assume for $n$, $19^{2^n}=1+m\cdot 2^{n+2}$, where $m$ is a positive odd integer. Therefore for $n+1$, we have $$19^{2^{n+1}}=(19^{2^n})^2=(1+m\cdot 2^{n+2})^2=1+2m\cdot 2^{n+2}+(m\cdot 2^{n+2})^2$$ $$\ \ \ \ \ \ \ \ \ \ \ \ \ =1+m\cdot 2^{n+3}+m^2\cdot 2^{2n+4}=1+(m+m^2\cdot 2^{n+1})\cdot 2^{(n+1)+2}$$

Because $m$ is odd, $m^2\cdot 2^{n+1}$ is even, therefore $m+m^2\cdot 2^{n+1}$ is odd. Therefore, we prove the case for $n+1$.

By induction, the problem is proved.