Shourya's function

First, we can simplify the number $14400$ into $30*480$ . Then, by using rule number 1, we can make such equation: $f(144000) = f(30) + f(480) - 1$ Then, since $480$ is $30*16$ , with using the first and third rule, we can write previous equation into $f(14400)=f(30)+(f(30)+f(16)-1)-1$ $=f(16)+2f(30)-2=f(16)+6$ $=2f(4)-1+6=4(2)-3+6=4f(2)+3$ Then, let's see the significance of the rule number 2. Let's assume that any prime number, $p$ , has a property that $f(p)=1$ Then, $f(p^{2})=2f(p)-1=1$ $f(p^{4}) = 2f(p^{2})-1=1...$ so the value of $f(p^{2n})$ will be always 1, when n is any positive integer. Therefore, this assumption breaks the second rule, so we can deduce that $f(p) \neq 1$ Then, let's go back to rule three. We can simplify $f(30)$ into such equation: $f(30) = f(6)+f(5)-1$ $=f(2)+f(3)+f(5)-2=4$ Then, $f(2)+f(3)+f(5)=6$ And since $2,3,5$ are all prime numbers, $f(2), f(3), f(5)$ all have to be greater than 1, so the only way is $f(2)=f(3)=f(5)=2$ Therefore, by plugging $f(2)=2$ to the previous equation, $f(14400)=8+3=\boxed{11}$

From the condition (1) we can obtain $f(1)=1$ by substituting $x=y=1$ .

For any positive integer $k>1$ . If $f(k)=1$ , then $f(k^2)=f(k)+f(k)-1=1$ .

Moreover $f(k^3)=f(k^2)+f(k)-1=1$ , and so on.

Which is contradict with the condition (2).

Thus there is only one positive integer $x$ satisfying $f(x)=1$ , that is $1$ .

Since $f(30)=4$ , by condition (1) we obtain $f(2)+f(3)+f(5)=6$ .

Because $f(2),f(3),f(5)$ are all positive integer more than $1$ , there is only one possible value of them that is $f(2)=f(3)=f(5)=2$ .

Now we move to computing $f(14400)$ .

By condition (1) we get $f(14400)=2f(30)+4f(2)-5=2(4)+4(2)-5=11$ .

From $f(30)=4$ ,

$f(14400)=f(30)+f(480)-1$ $=f(30)+(f(30)+f(16)-1)-1=6+f(16)$

$f(16)=f(4)+f(4)-1$ $=(f(2)+f(2)-1)+(f(2)+f(2)-1)-1=4f(2)-3$

Therefore, $f(14400)=3+4f(2)$

Since $f(30)=f(2)+f(3)+f(5)-2=4$ , $f(2)+f(3)+f(5)=6$

From the condition 2, none of the prime number can be equal to 1 because that will make its power become 1 and there will be infinitely many 1s. ( If $f(2)=1$ , then $f(4)=f(2)+f(2)-1=f(2)=1$ , $f(8)=f(2)+f(2)+f(2)-2=f(2)=1$ and so on)

So $f(2)=2$ .

So the value of $f(14400)=3+4f(2)=11$ .

First of all, we can easily prove that $f(1)=1$ . Now, let n be the greatest solution of $f(x)=1$ Then, $f(n^2)=2f(n)-1$ Therefore $n=1$ . Therefore, $f(2)>1, f(3)>1, f(5)>1$ Moreover, we have $f(30)=4=f(2)+f(3)+f(5)-2\Rightarrow f(2)=6-f(3)-f(5)$ yet $f(3)\ge 2, f(5)\ge 2$ so $f(2)\le 6-2-2=2 \land f(2)>1\Rightarrow f(2)=2$ We can now find $f(16)$ : $f(16)=f(2)+f(8)-1=2f(2)+f(4)-2=4f(2)-3=5$ Finally : $f(14400)=f(30)+f(480)-1=2f(30)+f(16)-2=8+5-2=\boxed{11}$

$f(14400)=f(900)+f(16)-1=2f(30)+f(16)-2=6+f(16)$

Also, $f(16)=2f(4)-1=2(2f(2)-1)-1=4f(2)-3$

We now know that $f(14400)=3+4f(2)$ . So if we can find $f(2)$ , the problem is solved.

If $f(p)=1$ for some prime $p$ , $f(p^{2})=2f(p)-1=1$ . Similarly, $f(p^{4})=1, f(p^{2^{n}})=1$ for all positive integers n.

But condition (2) states that $f(x)=1$ does not hold for infinitely many x, so $f(p)≥2$ . $f(30)=f(2)+f(15)-1=f(2)+f(3)+f(5)-2=4$ , so $f(2)+f(3)+f(5)=6$ . This means that $f(2)=f(3)=f(5)=2$ (since 2, 3, 5 are primes) otherwise they cannot add up to 6. So the answer is $3+4(2)=11$ .

The given condtion 1 imply for all positive integers $x_1, x_2 \cdots , x_n$ , $f(x_1 \cdot x_2 \cdots x_n)=f(x_1)+f(x_2x\cdot x_3 \cdots x_n)-1=\cdots = f(x_1)+f(x_2)+\cdots+f(x_n)-(n-1)$ , which we can have $f(14400)=f(2^6 \cdot 3^2 \cdot 5^2)=6f(2)+2f(3)+2f(5)-11$ . It suffices to determine the value of $f(2), f(3), f(5)$ .

Lemma: $f(n)\ge 2$ for all $n\neq 1$ .

If there exist $n\neq 1$ such that $f(n)=1$ , then $f(n^k)=f(n^{k-1})+f(n)-1=f(n^{k-1})=f(n^{k-2})=\cdots f(n)=1$ for all $k$ , condraticts condition 2.

Given $f(30)=4$ , then $f(2)+f(15)=5$ , $f(3)+f(10)=5$ , $f(5)+f(6)=5$ . This imply $f(2), f(3), f(5), f(6)$ are either 2 or 3. However, $f(2)+f(3)=f(6)-1$ , and $f(6)$ is either 2 or 3, we have $f(2)=f(3)=2$ . So $f(6)=3 \Rightarrow f(5)=2$ .

So $f(14400)=6f(2)+2f(3)+2f(5)-11=11$ .