Today I've encountered a question like The following;  If function $f$ satisfies $f(xy)=f(x)f(y)$ and $f(81)=3$ then find The value of $f(2)$?  What baffles me about this question is that I have to find The equation of the function in order to find $f(2)$ because $2$ is not a divisor of $81$ , using The property I found out that $f(3)=\sqrt[4]{3}$ and wondered if the function could be $f(x)=\sqrt[4]{x}$ (it satisfies the equation up there) but I do not know whether f gives $2$ a value like this or not. And there are many other functions that can be found.   So The question is how can İ get myself out of this ugly situation, and how can I find other $f$ functions that satisfy the constraints? What I am asking is not to prove that these functions are in type of $x^n$ I am trying to get what $f(2)$ is and see also whether this question is deficient ör cannot ve solved with The given details.  Thank you:)
35894
Let  $f(xy) =f(x)f(y)$ for all $x,y\geq 0$. Show that $f(x) = x^p$ for some $p$.    I am not very experienced with proof. If we let $g(x)=\log (f(x))$ then this is the same as  $g(xy) = g(x) + g(y)$   I looked up the hint and it says let $g(x) = \log f(a^x) $   The wikipedia page for functional equations only states the form of the solutions without proof.  Attempt  Using the hint (which was like pulling a rabbit out of the hat)  Restricting the codomain $f:(0,+\infty)\rightarrow (0,+\infty)$ so that we can define the real function $g(x) =  \log f(a^x)$ and we have $$g(x+y) = g(x)+ g(y)$$   i.e $g(x) = xg(1)$ as $g(x)$ is continuous (assuming $f$ is).  Letting $\log_a f(a) = p$ we get $f(a^x) =a^p $. I do not have a rigorous argument but I think I can conclude that $f(x) = x^p$ (please fill any holes or unspecified assumptions) Different solutions are invited
119668
My question is as follows: What methods can be used to find the set of functions $f:\mathbb{R}\to\mathbb{R}$ satisfying a certain functional equation. An example of a case where this applies is the following:  Find all functions $f:\mathbb{R}\to\mathbb{R}$ which satisfy the following equation: $$f\big(x^3\big)+f\big(y^3\big)=(x+y)\Big(f\big(x^2\big)+f\big(y^2\big)-f(xy)\Big):\forall x, y\in\mathbb R$$  I'm curious as to whether there are general methods (or strategies) for solving this type of question, or whether questions like these should just be handled on a case-by-case basis. Thanks in advance.
46840
1.find a for $f(x) = x^2 + 2(m − a)x + 3am −2 = 0$ such that for every m real, f has real roots 2.find m such that for every a real, f has real roots  My ideea is to demonstrate that $\delta=4(m-a)^2 - 4(3am-2) = 4(m^2-5am+a^2+2) &gt;0$ But I have no ideea how to do this.
79524
Find all continuous functions such that $f(x+y) = f(x)f(y)f(xy)$   I solved it and got the two very obvious solutions, $f(x) = 0$ and $f(x) = 1$. Any other such functions?
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Let $S = \{1,2,3,4\}$. Let $F$ be the sets of all functions from $S$ to $S$.  Now I think this statement is true:  $\forall f \in F , \exists g\in F$ so that $(g \circ f)(1) =2$  I suppose $f \in F$, and I need to show that $\exists g\in F$ so that $(g \circ f)(1) = 2$  Do I just make an example  of $g$ so that $(g \circ f)(1) = 2$ ?  Like $f(1) = y$  and then  $g (y)=2 ?
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The hint is: Let F has as elements $a_1,...,a_n$. Note that if $$f_i(x)=c(x-a_1)...(x-a_{i-1})(x-a_{i+1})...(x-a_n)$$, then $f_i(a_j)=0$ for $i\neq j$ and the value $f_i(a_j)$ can be controlled by the choice of $c\in F$. Use this to show that every function on F is a polynomial function.  Using Fraleigh’s seventh edition of A First Course in Abstract Algebra. Need help with problem 22.31c. I’m completely lost. Thank you!
83001
I am trying to solve the following functional equation, and could use some help.$$ 2f(x)=f(ax)$$  For some $a\in\mathbb{R}$. By repeated adding $2f(x)$ together we notice that $$2nf(x)=f(a^nx).$$  Also $$2mf(a^{-m}x)=f(x)\Rightarrow \frac{1}{2m}f(x)=f(a^{-m}x).$$  Putting it all together I have for all $m,n\in\mathbb{N}$, $$\frac{n}{m}f(x)=\frac{2n}{2m}f(x)=f(a^{n-m}x)$$  I am unsure of how to proceed from here.
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I have a function  $y = \sqrt{x^2 + 1}− x$ and I need to prove if it's a Injective function (one-to-one).  The function f is injective if and only if for all a and b in A, if f(a) = f(b), then a = b  $\sqrt{a^2 + 1} − a = \sqrt{b^2 + 1} − b$  $\sqrt{a^2 + 1} + b = \sqrt{b^2 + 1} + a$  $(\sqrt{a^2 + 1} + b)^2 = (\sqrt{b^2 + 1} + a)^2$  $a^2 + 1 + b^2 + 2b\sqrt{a^2 + 1} = b^2 + 1 + a^2 + 2a\sqrt{b^2 + 1}$  $2b\sqrt{a^2 + 1} = 2a\sqrt{b^2 + 1}$  $b\sqrt{a^2 + 1} = a\sqrt{b^2 + 1}$  $(b\sqrt{a^2 + 1})^2 = (a\sqrt{b^2 + 1})^2$  $b^2(a^2 + 1) = a^2(b^2 + 1)$  $b^2a^2 + b^2 = a^2b^2 + a^2$  $b^2 = a^2$  $\sqrt{b^2} = \sqrt{a^2}$  $b = \pm a$  then the function is not injective because $b$ in not equal to $a$, but using online  the function is injective.
94317
Find $f:f(xf(x)+f(y))=f(x)^2+y$  Domain and co-domain is real numbers I did the following:  Let $s=f(0)$  Then $f(f(y))=s^2+y$ so $f$ is surjective Also, $f(x)=f(y)\implies f(xf(x)+f(y))=f(xf(x)+f(x))\implies x=y$ so $f$ is injective So, $f$ is bijective. Letting $f(x)=0, y=0$  we get $f(f(0))=0 \implies s=f(0)=0$  In fact $f(f(x))=x$ for ALL $x$ Letting $x=1,y=0$ we get  $f(f(1))=f(1)^2\implies 1=f(1)^2 \implies f(1)=1$ or  $f(1)=-1$ If $f(1)=1$ we let $x=1$ to get  $f(y+1)=f(y)+1$  By induction this leads to $f(x)=x$ for all integers If $f(1)=-1$ we let $x=1$ to get  $f(y-1)=f(y)+1$  By induction we get $f(x)=-x$ for all integers How to extend the domain over real numbers I don't know, any help would be appreciated
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Find all functions $f:\mathbb R \to \mathbb R$ that satisfy $$ f(x)  + 3 f\left( \frac {x-1}{x} \right) = 7x $$ for all nonzero $x$.  My thoughts so far are to plug in $(x - 1)/x$ for $x,$ but it always comes out so complicated, so I don't really know where to go from there.
69405
If $x&gt;0$, $y&gt;0$, $z&gt;0$ and $x+y+z=6$ then find the minimum value of    $$ f(x,y,z)=\left(x+\frac{1}{y}\right)^2+\left(y+\frac{1}{z}\right)^2+\left(z+\frac{1}{x}\right)^2. $$   Thanks in advance.
120179
I'm having trouble with this functional equation, to find all functions $f:\mathbb R \to \mathbb R$ for which the following is valid for all $x,y\in \mathbb R$: $f(x) + f(x+f(y)) = y + f(f(x) + f(f(y)))$ I haven't been able to pinpoint any solution, nor to prove surjectivity or injectivity. So 0 progress. Tried with $x=y=0, x=0, y=0, x=-f(y), y=f(x)$, and even with some assumptions on bijectivity, but nothing helped. I don't know the source but seems like a pretty difficult problem, maybe even impossible to solve.
25349
Let $\mathbb{F}$ be a finite field.  Let $a_1,\dots,a_n \in \mathbb{F}$ be given.  I want to know whether there exists $x_1,\dots,x_n \in \mathbb{F}$ such that  $$a_1 x_1^2 + a_2 x_2^2 + \dots + a_n x_n^2 = 1.$$  My question: Can we characterize when this equation has a solution, i.e., by some criterion on $a_1,\dots,a_n$?  Is there a clean characterization, in terms of $a_1,\dots,a_n$ and the properties of $\mathbb{F}$?    Here's what I can see immediately.  First, if $n=1$, there is an easy answer (the equation has a solution iff $a_1$ is a square in $\mathbb{F}$).  Second, if any $a_i$ is a square in $\mathbb{F}$, the equation has a solution (set all other $x_j$'s to $0$), so we can assume that all the $a_i$'s are non-squares and that $n \ge 2$.  Third, if $-1$ is a square in $\mathbb{F}$, then the equation has a solution: we can take $x_2 =  \sqrt{-a_1/a_2} \cdot x_1$.  But what about the case where $-1$ is a non-square in $\mathbb{F}$?  Is it always guaranteed to have a solution in this case?
87297
Let $f(x) = x^2 + ax + b$ be a quadratic polynomial in which $a$ and $b$ are integers. If for a given $n$ , $f(n)f(n+1) = f(m)$ for some integer $m$ then the value of $m$ is ?  Do we have to substitute $x$ by $n$ and $n+1$ and then multiply it? or is there another method to get the solution because the mentioned process will give an extremely large equation.  Answer is $n(n+1)+an+b$
63194
If $f(x)=x+\int_{0}^{1}[xy^2+x^2y]f(y)dy$ where $x$ and$y$ are independent variable.Find $f(x).$  I tried to solve it.  $f(x)=x+\int_{0}^{1}[xy^2+x^2y]f(y)dy$  $f(x)=x+x\int_{0}^{1}y^2f(y)dy+x^2\int_{0}^{1}yf(y)dy$  I applied integration by parts but the expression did not simplify.Please guide me to reach the answer $f(x)=x+\frac{61}{119}x+\frac{80}{119}x^2.$
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Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that $$(x+y)(f(x)-f(y))=(x-y)f(x+y)$$  My attempt:   If $x=-y \not = 0$ then $0= 2x f(0)$ so $f(0)=0$.   Suppose for the sake of contradiction that $f(x)=f(x+\epsilon)$ for some $x$ and $\epsilon&gt;0$. Let $y=x+\epsilon$. Then $$0=\epsilon \cdot f(2x+\epsilon)$$ therefore as $2x+ \epsilon$ can take any real value $f$ is either strictly increasing or strictly decreasing or $f(x)=0 \; \; \forall \; \;x \in \mathbb{R}$.   Note that $f(x)=ax$ is a solution $\forall \; \;a \in \mathbb{R}$. Thanks so much for any help!
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Determine all $f: \mathbb{R} \rightarrow \mathbb{R}$ such that, for all $x, y \in \mathbb{R}$, $$f((x + 1) f(y)) = y (f(x) + 1).$$    Source: Vuong Lam Huy, on a Facebook group.