Problem 4

Call a rational number

r

powerful if

r

can be expressed in the form

\dfrac{p^k}{q}

for some relatively prime positive integers

p,q

and some integer

k>1

. Let

a,b,c

be positive rational numbers such that

abc=1

. Suppose there exist positive integers

x,y,z

such that

a^x+b^y+c^z

is an integer. Prove that

a,b,c

are all powerful.

Step 4 of 5: Force the equality

nz=m(y+z)

Previous step Next step

nz<m(y+z) \implies p^{nz}\Vert(\text{sum}),\ nz<nz+my \implies \text{contradiction} \implies nz=m(y+z)

Detailed analysis

The original divisibility also gives that $p^{nz+my}$ divides the sum $a_1^za_2^{y+z}b_1^x+b_1^{x+z}b_2^{y+z}$ . If $nz<m(y+z)$ strictly, the two terms of this sum have $p$ -adic valuations $nz$ and $m(y+z)$ with $nz$ strictly smaller, so the sum has valuation exactly $nz$ , which is less than $nz+my$ since $m>0$ — a contradiction. Hence $nz=m(y+z)$ .