I have the following problem: There are the Proth numbers $h\cdot 2^k+1$ with $h<2^k$ odd. Let $p$ be an odd prime and $r := \operatorname{ord}_p(2)$ be the multiplicative order of $2$ modulo $p$. Now I want to show that if $l$ is a natural number and $l\equiv k \pmod r$, then $h\cdot 2^k+1 \equiv 2^l+1\pmod r$.
Let $l\equiv k\pmod r \Longleftrightarrow r\mid k-l.$ Then we find a natural number $f$ with $f\cdot r = k-l$. Then $k = f\cdot r + l$. Now we have $$ h\cdot 2^k + 1 = h\cdot 2^{f\cdot r + l} +1= h\cdot (2^r)^f\cdot 2^l + 1. $$ And due to our choice of $r$ we have $$ h\cdot (2^r)^f\cdot 2^l + 1 \equiv h\cdot 2^l+1 \pmod r. $$
Did I forget something?
solution-verificationquestion to be on topic you must specify precisely which step in the proof you question, and why so. This site is not meant to be used as a proof checking machine. – Bill Dubuque Mar 19 '24 at 05:42