# Problem A. 479. (April 2009)

**A. 479.** Decide whether there exists a positive integer *n* that is divisible by 103 and satisfies .

Dutch competition problem, composed by *Hendrik Lenstra,* Leiden

(5 pont)

**Deadline expired on May 15, 2009.**

**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*(2*n*,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*(2*n*,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.

### Statistics:

11 students sent a solution. 5 points: Bodor Bertalan, Éles András, Nagy 235 János, Nagy 314 Dániel, Nagy 648 Donát, Somogyi Ákos, Tomon István, Tossenberger Anna, Varga 171 László, Weisz Ágoston. 4 points: Backhausz Tibor.

Problems in Mathematics of KöMaL, April 2009