# Is $\ 101\$ the only solution of $\ 2^{n-1}\equiv 203\mod n\$?

I wonder whether the congruence $$2^{n-1}\equiv 203\mod n$$ with integer $$n>1$$ has only the solution $$n=101$$.

Up to $$n=10^9$$, there is no solution.

Since $$n$$ must be odd, every prime factor $$p$$ of $$n$$ must have $$203$$ as a quadratic residue modulo $$p$$ and $$2^k\equiv 203 \mod p$$ must be solvable.

Deeper analysis reveals that the smallest possible prime factors are $$17$$ and $$53$$.

• If $$17$$ is a prime factor , $$n$$ must be of the form $$136k + 85$$.
• If $$53$$ is a prime factor , $$n$$ must be of the form $$2756k +477$$.
• how can you say that every prime factor must have $203$ as a quadratic residue modulo $p$? Could you elaborate? – vidyarthi Apr 5 at 7:35
• @vidyarthi Since $n$ is odd , $n-1$ is even , hence $2^{n-1}$ a perfect square. – Peter Apr 5 at 9:02
• Assuming $2^{n-1}\equiv 203\pmod n$, can you determine $n\mod 3$ or $n\mod 4$? – W-t-P Apr 6 at 8:56
• Is this question about $203$ and $n=101$ in particular, or is it really asking the general question, how to determine the greatest $n$ satisfying $2^{n-1} \equiv r$ (mod $n$), and/or whether a greatest $n$ always exists? If in fact there is a greatest $n$ for every given $r$, and the law of it can be even roughly determined, it might be unnecessary in the posted case to test values of $n$ beyond a limit far less than $10^9$. – Edward Porcella Apr 13 at 22:52
• @EdwardPorcella This question is only about $203$, the smallest case for which I could not find a solution, even with several tricks. I am actually interested in the general case, but the answer to this question is what Max answered. Since no further information was given by Max, I assume that he used more or less brute force. – Peter Apr 14 at 6:09

No. The next solution is $$n=33191065315201$$.