jmc

Setting $a=0$ and $b=0$ in the given functional equation, we get $$2f(0)=2f[(0){]}^2.$$Hence, $f(0)=0$ or $f(0)=1.$

Setting $a=0$ and $b=1$ in the given functional equation, we get $$2f(1)=[f(0){]}^2+[f(1){]}^2.$$If $f(0)=0,$ then $2f(1)=[f(1){]}^2,$ which means $f(1)=0$ or $f(1)=2.$ If $f(0)=1,$ then $[f(1){]}^2-2f(1)+1=[f(1)-1{]}^2=0,$ so $f(1)=1.$

We divide into cases accordingly, but before we do so, note that we can get to $f(25)$ with the following values: $$\begin{array}{rl}a=1,b=1:& 2f(2)=2[f(1){]}^2\Rightarrow f(2)=[f(1){]}^2\\ a=1,b=2:& 2f(5)=[f(1){]}^2+[f(2){]}^2\\ a=0,b=5:& 2f(25)=[f(0){]}^2+[f(5){]}^2\end{array}$$Case 1: $f(0)=0$ and $f(1)=0.$

From the equations above, $f(2)=[f(1){]}^2=0,$ $2f(5)=[f(1){]}^2+[f(2){]}^2=0$ so $f(5)=0,$ and $2f(25)=[f(0){]}^2+[f(5){]}^2=0,$ so $f(25)=0.$

Note that the function $f(n)=0$ satisfies the given functional equation, which shows that $f(25)$ can take on the value of 0.

Case 2: $f(0)=0$ and $f(1)=2.$

From the equations above, $f(2)=[f(1){]}^2=4,$ $2f(5)=[f(1){]}^2+[f(2){]}^2=20$ so $f(5)=10,$ and $2f(25)=[f(0){]}^2+[f(5){]}^2=100,$ so $f(25)=50.$

Note that the function $f(n)=2n$ satisfies the given functional equation, which shows that $f(25)$ can take on the value of 50.

Case 3: $f(0)=1$ and $f(1)=1.$

From the equations above, $f(2)=[f(1){]}^2=1,$ $2f(5)=[f(1){]}^2+[f(2){]}^2=2$ so $f(5)=1,$ and $2f(25)=[f(0){]}^2+[f(5){]}^2=2,$ so $f(25)=1.$

Note that the function $f(n)=1$ satisfies the given functional equation, which shows that $f(25)$ can take on the value of 1.

Hence, there are $n=3$ different possible values of $f(25),$ and their sum is $s=0+50+1=51,$ which gives a final answer of $n\times s=3\times 51=\boxed{153}$.