Problem 5

Determine all functions

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

such that

f(f(a)-b)+b\,f(2a)

is a perfect square for all integers

a

and

b

Step 4 of 5: Necessity: pin down f(0)

Previous step Next step

f(0)\in\{0,-1\};\qquad f(0)=-1\ \text{is impossible}

Detailed analysis

Let c denote f(0). Substitution at a=0 and b=c gives c+c²=z², and factoring the difference of squares yields c=0 or c=−1. If c=−1, substituting b=f(a) shows that f(a)f(2a)−1 is a square for every a. The standard prime-divisor argument for one more than a square forces every prime divisor of f(a) to be 2 or congruent to 1 modulo 4, with no factor 4. In particular this applies at −4. But substituting a=0 and b=3 makes f(−4)−3 a square, so f(−4) is congruent to 3 modulo 4, a contradiction. Hence f(0)=0.