MathLabs

Problem 1

Let Z+\mathbb{Z}^+ be the set of positive integers. Determine all functions f:Z+→Z+f:\mathbb{Z}^+\to\mathbb{Z}^+ such that a2+f(a)f(b)a^2+f(a)f(b) is divisible by f(a)+bf(a)+b for all positive integers aa and bb.
Step 2 of 6: Substitute a=1 for a lower bound
Previous stepNext step
b+1∣f(b)+1 ⟹ f(b)≥b for all bb+1\mid f(b)+1 \ \Longrightarrow\ f(b)\ge b\ \text{for all}\ b
Detailed analysis

Setting a=1a=1 gives b+1∣1+f(1)f(b)=1+f(b)b+1\mid1+f(1)f(b)=1+f(b). Writing f(b)+1=k(b+1)f(b)+1=k(b+1) for a positive integer kk, we get f(b)=k(b+1)−1≥(b+1)−1=bf(b)=k(b+1)-1\ge(b+1)-1=b. Hence f(b)≥bf(b)\ge b for every positive integer bb.