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 3 of 5: Valuation comparison at a prime p∣a1p\mid a_1
pn∥a1, pm∥b2,nz≤m(y+z),pnz∥a1za2y+zb1x, pm(y+z)∥b1x+zb2y+zp^n\Vert a_1,\ p^m\Vert b_2,\quad nz\le m(y+z),\quad p^{nz}\Vert a_1^za_2^{y+z}b_1^x,\ p^{m(y+z)}\Vert b_1^{x+z}b_2^{y+z}
Detailed analysis

Fix a prime p∣a1p\mid a_1 and let pn∥a1p^n\Vert a_1, pm∥b2p^m\Vert b_2 (so n,m≥1n,m\ge1). From a1z∣b2y+za_1^z\mid b_2^{y+z} we get nz≤m(y+z)nz\le m(y+z). Since gcd⁡(a1,b1)=gcd⁡(a2,b2)=1\gcd(a_1,b_1)=\gcd(a_2,b_2)=1, pp divides neither b1b_1 nor a2a_2, so pnz∥a1za2y+zb1xp^{nz}\Vert a_1^za_2^{y+z}b_1^x exactly while pm(y+z)∥b1x+zb2y+zp^{m(y+z)}\Vert b_1^{x+z}b_2^{y+z} exactly.