BdMO 2017 Dhaka divitional

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soyeb pervez jim
Posts: 21
Joined: Sat Jan 28, 2017 11:06 pm

BdMO 2017 Dhaka divitional

Unread post by soyeb pervez jim » Wed Mar 28, 2018 8:04 pm

For any rational numbers $x, y$ function $f(x)$ is a real number and $f(x+y)=f(x)f(y)-f(xy)+1$. Again $f(2017)\neq f(2018)$. $f(\frac{2017}{2018})=\frac{a}{b}$.Where $a,b$ are co-prime $a+b=?$

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samiul_samin
Posts: 1004
Joined: Sat Dec 09, 2017 1:32 pm

Re: BdMO 2017 Dhaka divitional

Unread post by samiul_samin » Sun Feb 10, 2019 3:59 pm

soyeb pervez jim wrote:
Wed Mar 28, 2018 8:04 pm
For any rational numbers $x, y$ function $f(x)$ is a real number and $f(x+y)=f(x)f(y)-f(xy)+1$. Again $f(2017)\neq f(2018)$. $f(\frac{2017}{2018})=\frac{a}{b}$.Where $a,b$ are co-prime $a+b=?$
Hint
$f(x)=x+1$
Answer
$6053$

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Arifa
Posts: 5
Joined: Wed Nov 22, 2017 8:09 pm

Re: BdMO 2017 Dhaka divitional

Unread post by Arifa » Fri Mar 22, 2019 11:41 am

samiul_samin wrote:
Sun Feb 10, 2019 3:59 pm
soyeb pervez jim wrote:
Wed Mar 28, 2018 8:04 pm
For any rational numbers $x, y$ function $f(x)$ is a real number and $f(x+y)=f(x)f(y)-f(xy)+1$. Again $f(2017)\neq f(2018)$. $f(\frac{2017}{2018})=\frac{a}{b}$.Where $a,b$ are co-prime $a+b=?$
Hint
$f(x)=x+1$
Answer
$6053$

হিন্ট টা কিভাবে আসলো?
Oops, I forgot

soyeb pervez jim
Posts: 21
Joined: Sat Jan 28, 2017 11:06 pm

Re: BdMO 2017 Dhaka divitional

Unread post by soyeb pervez jim » Sat Mar 23, 2019 3:53 pm

Original solution by Tonmoy.
Let $P(x,y)$ be the assertion $f(x+y)=f(x)f(y)-f(xy)+1$

$P(0, 0)$ $\Longrightarrow$ $f(0) = 1$.

Claim 1: $f(1) \neq 1$.
Proof of claim 1: If $f(1)=1$, then $P(x-1,1)$ $\Longrightarrow$ $f(x)=1$ $\forall x$. But it is given that $f(2017) \neq f(2018)$, a contradiction.

Claim 2: $f(-1) = 0$.
Proof of claim 2: $P(-1, 1)$ $\Longrightarrow$ $1 = f(0) = f(1)f(-1) - f(-1) + 1$.
So, $f(-1)(f(1) - 1) = 0$ $\Longrightarrow$ $f(-1) = 0$ since $f(1) \neq 1$ from the Claim $1$.

Then $P(x, -1)$ $\Longrightarrow$ $f(x - 1) = 1 - f(-x)$.....$(*)$

Claim 3: $f(1) \neq 0$.
Proof of claim 3: Let $f(1) = 0$. Then $P(1,1)$ $\Longrightarrow$ $f(2) = 1$.
Then $P \left(2, \dfrac {1} {2} \right)$ $\Longrightarrow$ $f \left(\dfrac {5} {2} \right) = f \left(\dfrac {1} {2} \right) + 1$.....$(i)$

Again, $P \left(1, \dfrac {1} {2} \right)$ $\Longrightarrow$ $f \left(\dfrac {3} {2} \right) = 1 - f \left(\dfrac {1} {2} \right)$.....$(ii)$

Now, $P \left(1, \dfrac {3} {2} \right)$ $\Longrightarrow$ $f \left(\dfrac {5} {2} \right) = 1- f \left(\dfrac {3} {2} \right) = 1- \left(1 - f \left(\dfrac {1} {2} \right) \right) = f \left(\dfrac {1} {2} \right)$, which clearly contradicts with $(i)$.

Now, Let $f(1) = a$.

$P(1, 1) \Longrightarrow f(2) = a^2 - a + 1$.
$P(2, 1) \Longrightarrow f(3) = a^3 - 2a^2 + 2a$.
$P(3, 1) \Longrightarrow f(4) = a^4 - 3a^3 + 4a^2 - 2a + 1$.
$P(2, 2) \Longrightarrow 2f(4) = f(2)^2 + 1$.

So, $2f(4) = f(2)^2 + 1 \Longrightarrow 2(a^4 - 3a^3 + 4a^2 - 2a + 1) = (a^2 - a + 1)^2 + 1$
$\Longrightarrow a(a-2)(a-1)^2 = 0$.

So, $a$ $\in$ ${0, 1, 2}$. But from Caim 1 and Claim 3, we know that $a \neq 1, a \neq 0$.

So, $f(1)=2$.
Now,
$P(x,1)$ $\Longrightarrow$ $f(x+1)=f(x)+1$ and so, by induction and using $(*)$, we get that $f(x+n)=f(x)+n$ and $f(n)=n+1$ for all integers $n$.

Then $P \left(q,\frac pq \right)$ $\Longrightarrow$ $q+f \left(\frac pq \right)=(q+1)f \left(\frac pq \right)-(p+1)+1$ and so $f \left(\frac pq \right)=\frac pq+1$

So, $f(x)=x+1$ $\forall x \in \mathbb{Q}$.

So, $f \left(\dfrac {2017} {2018} \right) = \dfrac {4035} {2018}$. So, $a + b = 6053$.

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