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Let $P(x,y)$ be the statement $f(x+y) + f(x)f(y) = f(xy) + f(x) + f(y)$.

$P(x,0) \implies f(x) + f(x)f(0) = f(0) + f(x) + f(0) \implies f(x)f(0) = 2f(0)$. If $f(0) \neq 0$, we can cancel it to give $f(x) = 2$ for all $x$, which is a solution. Otherwise, $f(0) = 0$.

$P(1,1) \implies f(2) + f(1)^2 = 3f(1)$. If $f(2) = 0$, this leads to $f(1) = 0$ or $f(1) = 3$. If $f(2) = 2$, this leads to $f(1) = 1$ or $f(1) = 2$.

$P(-1,1) \implies f(-1)f(1) = 2f(-1) + f(1)$. $f(1) = 2$ leads to a contradiction, so we are left with three cases. To recap, our current cases are $(f(1), f(2)) = (0,0), (1,2), (3,0)$.

$P(x,1) \implies f(x+1) + f(x)f(1) = 2f(x) + f(1)$. We can now apply induction (in both ways, going on the positive integers and the negative integers) to determine the value of $f$ on the integers:

Case 1:$f(1) = 0,$. Then $P(x,1) \implies f(x+1) = 2f(x)$. Inducting gives $f(x) = 0$ for all integer $x$.

Case 2:$f(1) = 1$. Then $P(x,1) \implies f(x+1) = f(x) + 1$. Inducting gives $f(x) = x$ for all integer $x$.

Case 3:$f(1) = 3$. Then $P(x,1) \implies f(x+1) = 3 - f(x)$. Inducting gives $f(x) = 3$ if $x$ is odd and $f(x) = 0$ if $x$ is even.

Let $x = \frac{p+1}{p}, y = p+1$ for nonzero integer $p$. Note that $x+y = xy$. Thus $P(x,y) \implies f(x)f(y) = f(x) + f(y)$. Since we know the value of $f(y)$, we can determine the value of $f(x)$. That is, the value of $f \left( 1 + \frac{p}{1} \right)$.

Using $P(x,1)$, we can induct again to determine the values of $f \left( n + \frac{1}{p} \right)$ for all integer $n$, including $n = 0$.

$P \left( \frac{1}{p}, n \right) \implies f \left( \frac{n}{p} \right) + f \left( \frac{1}{p} \right) f(n) = f \left( n + \frac{1}{p} \right) + f \left( \frac{1}{p} \right) + f(n)$. Since we know the values of $f \left( \frac{1}{p} \right)$, $f \left( n + \frac{1}{p} \right)$, and $f(n)$, we know the value of $f \left( \frac{n}{p} \right)$. That is, we know $f$ to the rationals. I haven't actually worked it out to figure out whether Case 3 still persists.

I'm not sure how to extend this to the reals. I'm also pretty sure this is too convoluted (that is, there should be a simpler solution). You might want to first try to solve the following: find all $f$ on the reals such that $f(x+y) = f(x) + f(y)$ and $f(xy) = f(x)f(y)$ for all reals $x,y$.

Hey should we make a thread for "Brilliant Functional Equation Contest" ? This is just an excuse for me to improve our functional equation as one problem in INMO is definitely on functional equations.

A functional equation problem shouldn't have the statement "if $f$ is a polynomial..."; that statement makes it a polynomial equation problem (which has quite a different method to solve them).

As mentioned by Ivan, I don't understand what you are doing here. I agree up the Case 2, but fail to follow form "Since the main functional equation ..."

The induction part is wrong, but it's a minor mistake; verifying that $f(x) = x$ works is simply plugging it into the equation. The problem is that other potential solutions haven't been ruled out.

You haven't shown that there is no other solution. In fact, as my solution suggests, if we restrict $f$ only to the integers, there exists another solution, and you haven't ruled it out.

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## Comments

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TopNewestLet $P(x,y)$ be the statement $f(x+y) + f(x)f(y) = f(xy) + f(x) + f(y)$.

$P(x,0) \implies f(x) + f(x)f(0) = f(0) + f(x) + f(0) \implies f(x)f(0) = 2f(0)$. If $f(0) \neq 0$, we can cancel it to give $f(x) = 2$ for all $x$, which is a solution. Otherwise, $f(0) = 0$.

$P(2,2) \implies f(4) + f(2)f(2) = f(4) + f(2) + f(2) \implies f(2)^2 = 2f(2)$. This gives $f(2) = 0$ or $f(2) = 2$.

$P(1,1) \implies f(2) + f(1)^2 = 3f(1)$. If $f(2) = 0$, this leads to $f(1) = 0$ or $f(1) = 3$. If $f(2) = 2$, this leads to $f(1) = 1$ or $f(1) = 2$.

$P(-1,1) \implies f(-1)f(1) = 2f(-1) + f(1)$. $f(1) = 2$ leads to a contradiction, so we are left with three cases. To recap, our current cases are $(f(1), f(2)) = (0,0), (1,2), (3,0)$.

$P(x,1) \implies f(x+1) + f(x)f(1) = 2f(x) + f(1)$. We can now apply induction (in both ways, going on the positive integers and the negative integers) to determine the value of $f$ on the integers:

Case 1:$f(1) = 0,$. Then $P(x,1) \implies f(x+1) = 2f(x)$. Inducting gives $f(x) = 0$ for all integer $x$.Case 2:$f(1) = 1$. Then $P(x,1) \implies f(x+1) = f(x) + 1$. Inducting gives $f(x) = x$ for all integer $x$.Case 3:$f(1) = 3$. Then $P(x,1) \implies f(x+1) = 3 - f(x)$. Inducting gives $f(x) = 3$ if $x$ is odd and $f(x) = 0$ if $x$ is even.Let $x = \frac{p+1}{p}, y = p+1$ for nonzero integer $p$. Note that $x+y = xy$. Thus $P(x,y) \implies f(x)f(y) = f(x) + f(y)$. Since we know the value of $f(y)$, we can determine the value of $f(x)$. That is, the value of $f \left( 1 + \frac{p}{1} \right)$.

Using $P(x,1)$, we can induct again to determine the values of $f \left( n + \frac{1}{p} \right)$ for all integer $n$, including $n = 0$.

$P \left( \frac{1}{p}, n \right) \implies f \left( \frac{n}{p} \right) + f \left( \frac{1}{p} \right) f(n) = f \left( n + \frac{1}{p} \right) + f \left( \frac{1}{p} \right) + f(n)$. Since we know the values of $f \left( \frac{1}{p} \right)$, $f \left( n + \frac{1}{p} \right)$, and $f(n)$, we know the value of $f \left( \frac{n}{p} \right)$. That is, we know $f$ to the rationals. I haven't actually worked it out to figure out whether Case 3 still persists.

I'm not sure how to extend this to the reals. I'm also pretty sure this is too convoluted (that is, there should be a simpler solution). You might want to first try to solve the following: find all $f$ on the reals such that $f(x+y) = f(x) + f(y)$ and $f(xy) = f(x)f(y)$ for all reals $x,y$.

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I believe that you're missing the constant solution $f(x) = 2$.

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I already wrote that one near the top.

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Can u too please post some functional equations to solve ? I am quite weak at that

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Try these:

$1$. If $f$ is a polynomial function satisying $2+f(x)f(y)=f(x)+f(y)+f(xy)$ for all real $x,y$ and if $f(2)=5$, find $f(f(2))$.

$2$. A polynomial function $f(x)$ satisfies the condition $f(x)f(\dfrac{1}{x})=f(x)+f(\dfrac{1}{x})$ if $f(12)=1729$, then find $f(10)$.

$3$.Find all functions $f$ from $R{0,1}$ to $R$ satisfying the functional relation $f(x)+f(\dfrac{1}{1-x})=\dfrac{2(1-2x)}{x(1-x)}$. Source:INMO

Buy the book Functional Equations by B.J. Venkatachala. It has very nice functional equations problem .

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Hey should we make a thread for "Brilliant Functional Equation Contest" ? This is just an excuse for me to improve our functional equation as one problem in INMO is definitely on functional equations.

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A functional equation problem shouldn't have the statement "if $f$ is a polynomial..."; that statement makes it a polynomial equation problem (which has quite a different method to solve them).

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Put $x=y=0$ $f(0)+f(0)^{2}=3f(0)$ This implies $f(0)=0$ or $f(0)=2$.

CaseI: $f(0)=2$ Now put $x=0$ $3f(y)=4+f(y)$ $\boxed{f(y)=2}$Case II: $f(0)=0$ Put $x=y=2$ $f(4)+f(2)^{2}=f(4)+2f(2)$ $f(2)=2$ or $f(2)=0$.Since the main functional equation all terms are only monic functions so another solution is $f(y)=0$.

Other solution is $f(y)=y$ which can be proved by induction.

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As mentioned by Ivan, I don't understand what you are doing here. I agree up the Case 2, but fail to follow form "Since the main functional equation ..."

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I mean that both sides have terms which have 1 as coefficient so each term can be zero.

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Yea Ivan has a point. I don't understand how u can use induction when the domain is reals ?

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The induction part is wrong, but it's a minor mistake; verifying that $f(x) = x$ works is simply plugging it into the equation. The problem is that other potential solutions haven't been ruled out.

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You haven't shown that there is no other solution. In fact, as my solution suggests, if we restrict $f$ only to the integers, there exists another solution, and you haven't ruled it out.

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