Problem A. 393. (February 2006)
A. 393. Prove that if n>1 is an integer such that 3n+4n is divisible by n then n is divisible by 7.
Deadline expired on March 16, 2006.
The number 3n+4n is odd and not divisible by 3, therefore n is odd as well and not divisible by 3.
Write n=2k+1. Then
3n+4n=3.9k+42k+13.(-1)k-1 (mod 5).
This number is not divisible by 5 for any value of n, so neither 5 is a divisor of n.
Let p be the smallest prime divisor of n and c is that residue class for which 4c3 (mod p). Then
4n(c2n-1)=(cn-1)((4c)n+4n)(cn-1)(3n+4n)0 (mod p)
|(1)||c2n1 (mod p).|
By Fermat's theorem,
Let d be the order of c modulo p. By (1) and (2), d is a common divisor of 2n and p-1, so d is a divisor of their GCD which is 2. Therefore d|2 and
If p|c-1 then 44c3 (mod p) which is a contradiction. So we have p|c+1 and
04(c+1)=4c+43+4=7 (mod p).
This implies p=7.
10 students sent a solution. 5 points: Erdélyi Márton, Estélyi István, Fischer Richárd, Gyenizse Gergő, Jankó Zsuzsanna, Kisfaludi-Bak Sándor, Nagy 224 Csaba, Paulin Roland, Tomon István. 1 point: 1 student.