Let p be a primes ,x,y,z be positive integers such that x<y<z<p and {x3p}={y3p}={z3p}. Prove that (x+y+z)|(x5+y5+z5).
Problem
Source: China Zhejiang Fuyang , 27 Jul 2014
Tags: modular arithmetic, number theory unsolved, number theory
17.08.2014 07:10
WLOG, we can just assume that x,y,z are distinct positive integers less than p. {x3p}={y3p}={z3p} means the following x^3\equiv y^3\equiv z^3 \pmod{p}. Thus, (x^{-1}y)^3\equiv (x^{-1}z)^3\equiv 1 \pmod{p}. Thus, for the primitive root g, we can say the following without loss of generality. x^{-1}y \equiv g^{\frac{p-1}{3}}, x^{-1}z \equiv g^{\frac{2(p-1)}{3}} \pmod{p}. Thus, y\equiv xg^{\frac{p-1}{3}}, z\equiv g^{\frac{2(p-1)}{3}} \pmod{p}. Meanwhile, considering (g^{\frac{p-1}{3}}-1)(g^{\frac{2(p-1)}{3}}+g^{\frac{p-1}{3}}+1)\equiv g^(p-1)-1\equiv 0 \pmod{p} and g^{\frac{p-1}{3}}\not\equiv 1 \pmod{p}, we can make the conclusion like this. x+y+z\equiv 0 \pmod {p} Thus, x+y+z=p _{or} 2p. Therefore, everything we have to do is to prove the following x^5+y^5+z^5\equiv 0 \pmod{p}. However, considering k^2+k+1 \mid k^{10}+k^5+1 it is easy to prove. Done
23.08.2014 22:33
It's exactly Poland 2003: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2070729&sid=64df69e2c903958785a9c858b780a3de#p2070729 just in the end you use that x^5+y^5+z^5\equiv x^3(x^2+y^2+z^2).
16.09.2024 11:51
We have x^3\equiv y^3\equiv z^3\pmod p. Note that p|y^3-x^3, and 0<y-x<p, so p|x^2+xy+y^2. Similarly, p|y^2+yz+z^2 and p|z^2+zx+x^2. We see p|x^2-y^2+xz-yz, so p|(x-y)(x+y+z). Hence, p|x+y+z. Therefore, x+y+z is either p or 2p. Note that p|x^2+y^2+z^2+2(xy+yz+zx) and p|2(x^2+y^2+z^2)+xy+yz+zx. Since p>3, we must have p|x^2+y^2+z^2. Hence, x^5+y^5+z^5\equiv x^3(x^2+y^2+z^2) \equiv 0 \pmod p. Also, x^5+y^5+z^5\equiv x+y+z\pmod 2. Therefore, x+y+z|x^5+y^5+z^5.