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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.

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.

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 ..."

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 $x,y$.

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).

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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## Comments

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TopNewestPut $x=y=0$ $f(0)+f(0)^{2}=3f(0)$ 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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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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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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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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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, $f(0) = 0$.

$P(2,2) \implies f(4) + f(2)f(2) = f(4) + f(2) + f(2) \implies f(2)^2 = 2f(2)$. 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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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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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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