Functional equation $f(n^2+1)-f(n^2)=1$

Question

Let $\mathbb{N}^*$ be the set of positive integers and $\mathcal{P}$ be the set of primes. I seek to prove that if $f:\mathbb{N}^∗\longrightarrow\mathbb{C}$ is completely multiplicative and such that : $$\forall n\in\mathbb{N}^*\,,\, f(n^2+1)-f(n^2)=1\,,$$ then $$ \forall p\in\mathcal{P},\ (p\equiv 1\;[4] \Rightarrow f(p)=p) \;\text{and}\; (p\equiv 3\;[4] \Rightarrow f(p)=\pm p)\,.$$ I came across this paper (see Theorem 2 (b2) p. 374), but I'm not really convinced by the end of the proof (see bottom of page 377). What do you think?

There are several things I don't understand:

• Isn't the second case rather $N=Q$ with $Q≡3\;[4]$ (not $N=Q^2$ )?
• It seems to me that all the prime factors $\pi$ of $\frac{Q^2+1}{2}$ are such that $π≡1\;[4]$, right?
• If we correct the induction hypothesis by "...if $p<N$, $q^2<N$ ...", a new problem arises : why would we have $f(n^2)=n^2$ ? How can we be sure that a prime factor $\pi$ of $n$ such that $\pi\equiv3\;[4]$ verifies $\pi^2<N$ ?

(Edit) I've opened another thread on this subject.

Answer 1

Let's consider the case where n=1n=1: We have f(12+1)−f(12)=f(2)−f(1)=1f(12+1)−f(12)=f(2)−f(1)=1.

Next, consider the case where n=2n=2: We have f(22+1)−f(22)=f(5)−f(4)=1f(22+1)−f(22)=f(5)−f(4)=1.

From these two cases, we can observe that the difference between the values of ff for consecutive perfect squares is always 1.

In other words, for any positive integer mm, we have f((m+1)2)−f(m2)=1f((m+1)2)−f(m2)=1.

We can express this relationship recursively as follows: f((m+1)2)=f(m2)+1f((m+1)2)=f(m2)+1.

We can start with an initial condition f(1)=cf(1)=c, where cc is a constant.

Using the recursive relationship, we can compute the values of ff for consecutive perfect squares: f(2)=f(12)+1=c+1f(32)=f(22)+1=(c+1)+1=c+2f(42)=f(32)+1=(c+2)+1=c+3f(52)=f(42)+1=(c+3)+1=c+4 f(2)f(32)f(42)f(52)​=f(12)+1=c+1=f(22)+1=(c+1)+1=c+2=f(32)+1=(c+2)+1=c+3=f(42)+1=(c+3)+1=c+4​

We can observe that the values of ff for consecutive perfect squares form an arithmetic sequence with the common difference of 1 and the initial term cc.

Therefore, the solution to the given functional equation is f(n)=n+c−1f(n)=n+c−1, where cc is any constant.