Problem 5

Let

f

be a function from the set of integers to the set of positive integers. Suppose that, for any two integers

m

and

n

, the difference

f(m)-f(n)

is divisible by

f(m-n)

. Prove that, for all integers

m

and

n

with

f(m)\le f(n)

, the number

f(n)

is divisible by

f(m)

Step 1 of 5: Every value divides f(0)

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In plain words

Comparing any point to itself minus zero shows f at that point already divides its own gap to f(0), and a number always divides itself, so it divides f(0) too.

f(x)\mid f(0)

Detailed analysis

Setting $n=0$ in the hypothesis gives $f(x)\mid f(x)-f(0)$ for every integer $x$ . Since $f(x)$ always divides itself, it also divides the difference $f(x)-\big(f(x)-f(0)\big)=f(0)$ ; that is, $f(x)\mid f(0)$ for every integer $x$ .