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You're right.   I made an incorrect observation.

The idea is that the assumption labeled (*) in the proof completely specifies the recurrence.  The base case is when n=0.  We just have to work out the possible values of f of the base case, and the necessary connections from the base case to the case of n=25.  Using (*), the base case value of f is governed by

    2f(0) = 2f(0)^2

 and the connections leading from 0 to 25 are governed by the conditions
    2f(1) = f(1)^2 + f(0)^2
    f(2) = f(1)^2
    2f(5) = f(2)^2+f(1)^2
    2f(25) = f(5)^2 + f(0)^2

Using these connections and working buttom up, we find all the possible values for f(0), f(1), f(2), f(5) and f(25).

No need to thank.   I'm simply enjoying myself.
 

Let $d$ be the distance.
Let $t_1$ be the time it takes to cover this distance at the rate $r_1$.
Let $t_2$ be the time it takes to cover this distance at the rate $r_2$.

Define $t_a$ to be the A.M. of the two times, that is, $t_a := (t_1+t_2)/2$.
Define $r_a$ to be the H.M. of the two rates, that is, $r_a := 2/((1/r_1)+(1/r_2))$.

Theorem. If one wants to cover distance $d$ in $t_a$ time, one should go at
the rate $r_a$.    Conversely, if one covers the distance at the rate $r_a$, one
will use up $t_a$ time.

Proof.   It's enough to show that $t_a\cdot r_a = d$.   We have
\begin{eqnarray*}
t_a\cdot r_a
&=& \dfrac{t_1+t_2}{2}\cdot \dfrac{2}{\dfrac{1}{r_1} + \dfrac{1}{r_2}} \\
&=& \dfrac{t_1+t_2}{2}\cdot \dfrac{2}{\dfrac{t_1}{d} + \dfrac{t_2}{d}} \\
&=& \dfrac{t_1+t_2}{2}\cdot \dfrac{2}{\dfrac{t_1+t_2}{d}} \\
&=& \dfrac{t_1+t_2}{2}\cdot \dfrac{2d}{t_1+t_2} \\
&=& d.
\end{eqnarray*}

The theorem extends to any positive integer $n$.

Let $d$ be the distance.
Let $t_1$ be the time it takes to cover this distance at the rate $r_1$.
Let $t_2$ be the time it takes to cover this distance at the rate $r_2$.
...
Let $t_n$ be the time it takes to cover this distance at the rate $r_n$.

Define $t_a$ to be the A.M. of the $n$ times.
Define $r_a$ to be the H.M. of the $n$ rates.

Theorem. If one wants to cover distance $d$ in $t_a$ time, one should go at
the rate $r_a$.    Conversely, if one covers the distance at the rate $r_a$, one
will use up $t_a$ time.

With this theorem, one can pose some seemingly more complicated questions like this one

Question: Dr. Worm leaves his house at exactly 7:20 a.m. every morning.
When he averages 30 miles per hour, he arrives at his workplace ten minutes late.  
When he averages 40 miles per hour, he arrives at his workplace five minutes late.  
When he averages 60 miles per hour, he arrives fifteen minutes early.
What speed should Dr. Worm average to arrive at his workplace precisely on time?
 

Convince yourself that the decimal representation of $n/30$ terminates if and only if $n$ is divisible by 3.   If you believe this, then counting those $n$ between 1 and 474 is a simple matter.

Alternatively
\[
8a+4b+2c+d
=a\cdot 2^2 + b\cdot 2^2 + c\cdot 2+d
= (3\cdot 2^2-5\cdot 2 +4)(7-2)
= 6\cdot 5
= 30.
\]
 

We can write 44 as a linear combination of $3n+4$ and $n+16$ like this
\[44=-(3n+4)+3(n+16)\]
so 44 must be a multiple of $\gcd(3n+4, n+16)$.  Since $44=2^2\cdot 11$, so $\gcd(3n+4, n+16)$ can only be one of 1, 2, 4, 11, 22, and 44.  I'll show in the following table that any of these six numbers can actually be the gcd of $3n+4$ and $n+16$ for some positive integer $n$.

\[\begin{array}{cccc}
    n & 3n+4 & n+16 & \gcd(3n+4, n+16) \\ \hline
    1 & 7 & 17 & 1 \\
    2 & 10 & 18 & 2 \\
    4 & 16 & 20 & 4 \\
    17 & 55 & 33 & 11 \\
    6 & 22 & 22 & 22 \\
    28 & 88 & 44 & 44
\end{array}\]

Thus, the answer is $1+2+4+11+22+44=84$.

Write $t=(a+b)(a+c)(a+d)(b+c)(b+d)(c+d)$.  I claim that the minimum value of $|t|$ is 0.   It's enough to come up with four distinct complex numbers $a$, $b$, $c$, $d$ such that $|a|=|b|=|c|=|d|$ and $a+b+c+d=0$, with $|t|=0$.  This is because the absolute value of a complex number is nonnegative, so 0 is the least value possible.   I now take $a=1$, $b=-1$, $c=i$, and $d=-i$.  It's easy to see that all hypotheses are satisfied, and $|t|=0$ because $a+b=0$, and $a+b$ is one of the factor of $t$.

Let's focus on the letter O's.   First note that the middle O cannot participate in any valid ROOM path, so we may as well get rid of it.   Now create a graph by letting each letter O be a vertex, and joining two neighboring vertices if we can go from one O to another by taking one step in any direction (up, down, left, right, or diagonal).

Here is the picture of the graph where each o is a vertex.  The edges are represented by the sequence of dashes or vertical line.   We also name the edges by the letters from A to H.

          A      B
      o-----o-----o
   C |                  | D
      o                 o
   E |                  | F
      o-----o-----o
          G     H
 

Now, to each vertex let's write down the number of the letters R and M it can go to in one step.   For example, entry 3r,2m means the vertex at that position can go to 3 different R's in one step, and 2 different M's in one step, etc.

    3r,2m 3r,0m 3r,2m
    0r,3m           0r,3m
    3r,2m 3r,0m 3r,2m
 

Now, to each edge we count how many ways it can be part of a ROOM path.  It is part of a ROOM path iff one end vertex is connected to R and the other end vertex is connected to M.   We show the calculation below.

    A: 3 * 2 = 6
    B: 3 * 2 = 6
    C: 3 * 3 = 9
    D: 3 * 3 = 9
    E: 3 * 3 = 9
    F: 3 * 3 = 9
    G: 3 * 2 = 6
    H: 3 * 2 = 6
 

So the total number of ways equals $6+6+9+9+9+9+6+6 = 60$.
 

See my answer at your original post https://web2.0calc.com/questions/hey-guys-really-need-help-hints-would-even-help-really-stuck#r10

Everything depends on this given assumption
\[
    2f(a^2+b^2) = [f(a)]^2 + [f(b)]^2\qquad\qquad(*)
\]
for all nonnegative integers $a$ and $b$.

Letting $b=0$ in $(*)$ we have another general formula
\[
    2f(a^2) = f(a)^2+f(0)^2\qquad\qquad\qquad(**)
\]
for all nonnegative integer $a$.

Letting $a=b=0$ in $(*)$ we get 
\[
    2f(0) = 2f(0)^2.
\]
Solving for $f(0)$ we get $f(0) = 0, 1$.

Letting $a=1$ in $(**)$ we get 
\[
    2f(1)=f(1)^2+f(0)^2.
\]
When $f(0)=0$, we solve for $f(1)$ to get $f(1) = 0, 2$;
when $f(0)=1$, we solve for $f(1)$ to get $f(1)=1$.

The current situation is like this

   | f(0) | 0 | 0 | 1 |
   | f(1) | 0 | 2 | 1 |

Setting $a = b = 1$ in $(*)$ we get 
\[
    f(2) = f(1)^2
\]
This implies that $f(2) = 0$ when $f(1) = 0$;
$f(2) = 4$ when $f(1) = 2$;
and $f(2) = 1$ when $f(1) = 1$;

The current situation is like this

 
   | f(0) | 0 | 0 | 1 |
   | f(1) | 0 | 2 | 1 |
   | f(2) | 0 | 4 | 1 |
 

Setting $a=2$ and $b=1$ in $(*)$ we have
\[
    2f(5) = f(2)^2 + f(1)^2
\]

When $f(1)=f(2)=0$, we solve for $f(5)$ to get $f(5) = 0$.
When $f(1)=2$ and $f(2)=4$, we solve for $f(5)$ to get $f(5)=10$.
When $f(1)=f(2)=1$, we solve for $f(5)$ to get $f(5)= 1$.

The current situation is like this

   | f(0) | 0 |   0 | 1 |
   | f(1) | 0 |   2 | 1 |
   | f(2) | 0 |   4 | 1 |
   | f(5) | 0 | 10 | 1 |
 

Letting $a=5$ in $(**)$ we get
\[
    2f(25) = f(5)^2 + f(0)^2
\]

When $f(0)=f(5)=0$, we solve for $f(25)$ to get $f(25) = 0$.
When $f(0)=0$ and $f(5)=10$, we solve for $f(25)$ to get $f(25)=50$.
When $f(0)=f(5)=1$, we solve for $f(25)$ to get $f(25)= 1$.

The final situation is like this

   | f(0)  | 0 |   0 |  1 |
   | f(1)  | 0 |   2 |  1 |
   | f(2)  | 0 |   4 |  1 |
   | f(5)  | 0 | 10 |  1 |
   | f(25) | 0 | 50 | 1 |

So $n = 3$ and $s=0+50+1=51$, and $n\times s = 3\times 51 = 153$.