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

Question

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 $2756k +477$.

Answer 1

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