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 3 of 6: Substitute b=1 for an upper bound

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f(a)+1\mid a^2-1 \ \Longrightarrow\ f(a)\le a^2-2\ \text{for}\ a\ge2

Detailed analysis

Setting $b=1$ gives $f(a)+1\mid a^2+f(a)$ . Since $a^2+f(a)=\big(a^2-1\big)+\big(f(a)+1\big)$ , this is equivalent to $f(a)+1\mid a^2-1$ . For $a\ge2$ , $a^2-1\ge3>0$ , so the divisor $f(a)+1$ is at most $a^2-1$ , giving $f(a)\le a^2-2$ .