Show that $f^k(m)$ contains a perfect square

Question

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1. At one point, it says, Thus it is enough to assume $m\in A$. Is that correct? I think they meant $f(m)$ instead of $m$.
2. Secondly, there's another error where they claim $\lfloor f(m) \rfloor =k$. It's supposed to be $\left\lfloor \sqrt f(m) \right\rfloor =k$

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DOUBTS :

1. In the very last line they say This clearly implies that $f^{2j}= (k+j)^2$ but I don't find it quite trivial. Can "you" please explain this part?

I think that the floored part is $k$, then, $k+1$, then $k+2$, $\cdots$. But then $f^{2j}$ doesn't come similar to theirs.

Answer 1

They do mean the $f(m)\in A$, and this is unclear because of the poor choice of notation because in the definition of A and B, the author uses $m$ as an element of the sets, but $m$ had already been used as the starting number.

I believe you are right about your second error.

You can notice that the remainder of $f^2(m)$ that is not a square is $j-1$ which is one less than $j$, and $k$ increased by 1 to $k+1$. The idea is that every time 2 compositions of $f$ is applied, the remainder will decrease by 1 and the square will increase by 1, so $f^{2j}(m) = (k+j)^2 + 0$.

Answer 2

They mean $m\in A.$ What they are saying is that it is enough to prove the theorem for numbers in $A.$ To do this, they reduce the case of a number in $B$ to the case of a number in $A$. So what they say is, "Suppose $m\in B.$ Then either $f(m)$ is a square, or $f(m)\in A.$ Now if the theorem is true for numbers in A then we know that one of $f^2(m), f^3(m),\dots$ is a square." Obviously if this is true, then one of $m, f(m), f^2(m), f^3(m),\dots$ is a square, so if we prove it for numbers in $A,$ it's also true for numbers in $B.$

This is too long-winded, but I can't seem to express it compactly at the moment.