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 6 of 6: Infinitely many divisors force the value

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b\big(f(b)-b\big)=0\ \Longrightarrow\ f(b)=b\ \text{for every}\ b\in\mathbb Z^+

Detailed analysis

For fixed $b$ , step 5 shows $p+b$ divides the fixed integer $b(f(b)-b)$ for every odd prime $p$ . Since $p+b$ can be made arbitrarily large while $b(f(b)-b)$ does not depend on $p$ , and a nonzero integer has only finitely many divisors, this forces $b(f(b)-b)=0$ . As $b>0$ , $f(b)=b$ . This holds for every positive integer $b$ , so $f(n)=n$ for all $n$ , and this function clearly satisfies the original condition since $f(a)+b=a+b$ divides $a^2+ab=a(a+b)$ .