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Let $\Phi $ be the set of functions $f:\mathbb{N}\rightarrow \mathbb{N}$, with $f(a^{2}-b^{2})=f(a)^{2}-f(b)^{2},\forall a,b\in \mathbb{N},a\geq b$.
a) Determine $S=\left \{ f(1)|f\in \Phi \right \}$.
b) Prove that $\Phi$ has two elements.

What I've done so far:
a)If we put $b=a$ we get $f(0)=0$.
If we put $b=0$, then $f(a^2)=f(a)^2$. Replacing $a$ by $1$, $f(1)=f(1)^2$, hence $S=\left \{ 0,1 \right \}$.
b)$f(a)^{2}-f(b)^{2}=f(a^{2}-b^{2})\geq 0$, so $f(a)^{2}\geq(b)^{2}$, resulting that $f$ is increasing.
Also, if $f(1)=0$, $f(4)=f(2^{2})=f(2)^{2}=f(2)^{2}-f(1)^{2}=f(2^{2}-1^{2})=f(3)$, so $0=f(4)^2-f(3)^2=f(4^2-3^2)=f(7)$.

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    You need to give an example of such f with f(1)=0. Also I assume you have $0\in \mathbb{N}$? – 2017-02-07
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    Hint: $\Phi$ contains both $f(x)=x$ and $f(x)=0$, so you would want to prove it has no others. – 2017-02-07
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    I found that $f(7)=0$ if $f(1)=0$, but I don't know how to prove that $f=0$. – 2017-02-07

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Partial answer

We know $f(a^2)=f^2(a)$, $f$ is increasing and $f(1) \in \{0,1\}$.

  1. Suppose $f(1)=0$. Then $f(7)=0$ as OP proved, therefore $f(n)=0, \forall n \le 7$. We'll use induction to prove $f(n)=0, \forall n$. Suppose $f(n)=0, \forall n \le k$. Because $k \lt k + 1 \le k^2$ it follows $f(k+1) \le f(k^2) = f^2(k)=0$
  2. Suppose $f(1)=1$. We have to prove $f(n)=n, \forall n$