MathLabs

Problem 4

Call a rational number rr powerful if rr can be expressed in the form pkq\dfrac{p^k}{q} for some relatively prime positive integers p,qp,q and some integer k>1k>1. Let a,b,ca,b,c be positive rational numbers such that abc=1abc=1. Suppose there exist positive integers x,y,zx,y,z such that ax+by+cza^x+b^y+c^z is an integer. Prove that a,b,ca,b,c are all powerful.
Step 5 of 5: Conclude a1a_1, hence aa, is powerful
k:=y+zgcd⁡(z,y+z)>1,k∣n ∀p∣a1  ⟹  a1=tk, gcd⁡(t,b1)=1  ⟹  a=tkb1 is powerfulk:=\frac{y+z}{\gcd(z,y+z)}>1,\quad k\mid n\ \forall p\mid a_1\implies a_1=t^k,\ \gcd(t,b_1)=1\implies a=\frac{t^k}{b_1}\text{ is powerful}
Detailed analysis

From nz=m(y+z)nz=m(y+z) with n,mn,m positive integers, dividing through by gcd⁡(z,y+z)\gcd(z,y+z) shows nn must be divisible by k:=(y+z)/gcd⁡(z,y+z)k:=(y+z)/\gcd(z,y+z), and k>1k>1 because z<y+zz<y+z. Since this holds for every prime pp dividing a1a_1, every exponent in the prime factorization of a1a_1 is divisible by kk, so a1=tka_1=t^k for some positive integer tt. Because gcd⁡(a1,b1)=1\gcd(a_1,b_1)=1, also gcd⁡(t,b1)=1\gcd(t,b_1)=1, and thus a=tk/b1a=t^k/b_1 has the required form. By the symmetric argument (with the corresponding variables exchanged), bb and cc are also powerful.