For a positive integer $n>1$, let $g(n)$ denote the largest positive proper divisor of $n$ and $f(n)=n-g(n)$. For example, $g(10)=5$, $f(10)=5$ and $g(13)=1$, $f(13)=12$. Let $N$ be the smallest positive integer such that $f(f(f(N)))=97$. Find the largest integer not exceeding $\sqrt{N}$.

Step 1: Let $p$ be the smallest prime factor of $x$. Then $g(x)=x/p$, so $f(x)=x(1-1/p)$.

Step 2: If $f(x)=k$ where $k$ is a prime number:

• If $x$ is prime, then $f(x)=x-1=k⟹x=k+1$ (which is even and composite if $k\ge 3$).

• If $x$ is composite, the smallest prime factor $p$ must satisfy $x(1-1/p)=k$. For $x$ to be minimized, we set $p=2$, which gives $x/2=k⟹x=2k$.

Step 3: We want to solve $f(f(f(N)))=97$ for the smallest $N$.
Let $Y=f(f(N))$. Since $f(Y)=97$ and $97$ is prime, the smallest possible value for $Y$ is:

Step 4: Now solve $f(x)=194$ where $x=f(N)$:

• If $x$ is composite and $p=2$: $x/2=194⟹x=388$.

• If $x$ is composite and $p=3$: $x(2/3)=194⟹x=291$. The smallest prime factor of 291 is indeed 3. This is smaller than 388.

Thus, the smallest value for $x$ is $291$, so $f(N)=291$ or $f(N)=388$.

Step 5: Now solve $f(N)=388$:
Since $388$ is composite, if $N$ is prime, then $f(N)=N-1=388⟹N=389$. Since 389 is prime, this is a valid solution! Let's check $f(N)=291$:

• If $N$ is prime, $N-1=291⟹N=292$ (composite, invalid).

• If $N$ is composite and $p=2$, $N/2=291⟹N=582>389$.

Step 6: Therefore, the smallest positive integer is $N=389$.

Step 7: Finally, find the largest integer not exceeding $\sqrt{N}$: