Solution. Suppose such an n exists. Since n is divisible by 103 we conclude , and since 103 is odd we conclude . Furthermore 103 is prime, and Fermat's theorem yields . For d=gcd(2n,102) this yields . Note that 102=6×17. If n is not divisible by 17, then d must be a divisor of 6, and would imply ; a contradiction. Hence n is divisible by 17.
The original congruence yields , and since 17 is odd we conclude . Furthermore 17 is prime, and little Fermat yields . For e=gcd(2n,16) this yields . The cases e=1, e=2, e=4 yield immediate contradictions. Hence e is divisible by 8, and n is divisible by 4.
The original congruence yields . But now the LHS is divisible by 4, whereas the RHS is not. This contradiction shows that no such n exists.