Problem 1

Let

\mathbb{Z}^+

be the set of positive integers. Determine all functions

f:\mathbb{Z}^+\to\mathbb{Z}^+

such that

a^2+f(a)f(b)

is divisible by

f(a)+b

for all positive integers

a

and

b

Step 4 of 6: Nail down the value at odd primes

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2f(p)\mid p^2+f(p)f(f(p))\ \Longrightarrow\ f(p)\mid p^2\ \Longrightarrow\ f(p)=p\ \text{for every odd prime}\ p

Detailed analysis

For an odd prime $p$ , set $a=p$ , $b=f(p)$ : the condition reads $f(p)+f(p)\mid p^2+f(p)f(f(p))$ , i.e. $2f(p)\mid p^2+f(p)f(f(p))$ . Since $f(p)\mid f(p)f(f(p))$ , also $f(p)\mid p^2+f(p)f(f(p))$ , hence $f(p)\mid p^2$ . As $p$ is prime, $f(p)\in\{1,p,p^2\}$ ; step 2 excludes $1$ (since $f(p)\ge p>1$ ) and step 3 excludes $p^2$ (since $f(p)\le p^2-2$ ). Hence $f(p)=p$ .