epitomy01 wrote: Find all functions $ f: \mathbb{Q}^{ + } \mapsto \mathbb{Q}^{ + }$ such that: $ f(x) + f(y) + 2xy f(xy) = \frac {f(xy)}{f(x + y)}$. 1) When $ x = y = 1$ thus $ f(2) = \frac {1}{4}$ $ x = y = 2$ then $ f(4) = \frac {1}{16}$ $ x = 1;y = 2$ then $ f(1) + \frac {1}{4} + 1 = \frac {1}{4f(3)}$ $ x = 1;y = 3$ then $ f(3) = \frac {f(1)}{9}$. Sice the third and fourth equation we see that $ f(1) = 1$. $ x\in \mathbb N^*;y = 1$ then $ \frac {1}{f(x + 1)} = 1 + \frac {1}{f(x)} + 2x$ By induction (very easily) we have - $ f(n) = \frac {1}{n^2}\forall n \in \mathbb N^*$ - $ \frac {1}{f(x + k)} = k^2 + 2kx + \frac {1}{f(x)}\forall x\in \mathbb{Q}^{ + }; k\in \mathbb N^*$. When $ x = \frac {1}{k}; y = k$ ($ k\in \mathbb N^*$) the seventh equation shows us $ f(\frac {1}{k}) = k^2\forall k\in \mathbb N^*$. Replace $ x$ by $ \frac {1}{q}$ ($ q\in \mathbb N^*$), the sixth implies$ f(x + \frac {1}{q}) = \frac {1}{(x + \frac {1}{q})^2} \forall x\in \mathbb Q^ + ; q\in \mathbb N^*$). Let $ a\in \mathbb Q^ +$ then $ a = \frac {p}{q}$ with $ p;q\in \mathbb N^*$. In the first equation, replace $ x = p;y = \frac {1}{q}$ we will have $ f(p) + f(\frac {1}{q}) + 2\frac {p}{q}f(a) = \frac {f(a)}{f(p + \frac {1}{q})}$ It means $ f(a) = \frac {q^2}{p^2}$. In conclusion, $ f(x) = \frac {1}{x^2}\forall x\in \mathbb Q^ +$. Done!

Corect solution .

I have another way.After I got the answer for postive integer n,I let x=p/q,y=q and then let x=p/q,y=2q and then x=(p/q)+q,y=q,and solved the equation.

Standard. Denote $P(x,y)$ as the assertion that $f(x)+f(y)+2xyf(xy)=\frac{f(xy)}{f(x+y)}$. $P(1,1)$ gives $f(2)=\frac{1}{4}$. $P(n,1)$ gives $f(n+1)=\frac{f(n)}{(2n+1)f(n)+f(1)}$, so $f(3)=\frac{1}{5+4f(1)}$ and $f(4)=\frac{1}{7+5f(1)+4f(1)^2}$. $P(2,2)$ gives $f(4)=\frac{1}{16}$, so we have $7+5f(1)+4f(1)^2=16$, giving $f(1)=1$. Now $P(n,1)$ gives $f(n+1)=\frac{f(n)}{(2n+1)f(n)+1}$. By induction on $n$ we can easily get that $f(n)=\frac{1}{n^2}$ for all $n \in \mathbb{N}$.

Now I claim that $k^2f(kx)=f(x)$ for all $k \in \mathbb{N}$ by induction. Perform $P(x,k)$, and this gives $$f(x+k)=\frac{f(x)}{(k^2+2kx)f(x)+1}$$Perform $P(x+1,k)$, with (0), we get $$f(x+1+k) = \frac{f(x+1)}{(k^2+2kx+2k)f(x+1)+1}=\frac{f(x)}{(k^2+2kx+2k+2x+1)f(x)+1}$$Perform $P(x,k+1)$, and this gives $$f(x+k+1) = \frac{(k+1)^2f((k+1)x)}{(k+1)^2f(x)+2x(k+1)^3f((k+1)x)+1}$$ Comparing these two, we get $$(k+1)^2f(x)^2+f(x) = (k+1)^4f((k+1)x)+(k+1)^2f((k+1)x)$$which is enough to say that $f((k+1)x)=\frac{f(x)}{(k+1)^2}$. Now we have $f(nx)=\frac{f(x)}{n^2}$ for all $n \in \mathbb{N}$. We now have $\frac{1}{p^2}=f(p)=f(\frac{p}{q} \cdot q) = \frac{f(\frac{p}{q})}{q^2}$, giving $f(\frac{p}{q}) = \frac{q^2}{p^2}$, giving $f(x)=\frac{1}{x^2}$.

Let $p,q,n$ be arbitrary natural numbers. $P(x,1)\Rightarrow f(x+1)=\frac{f(x)}{f(x)+2xf(x)+f(1)}$ Since $P(1,1)\Rightarrow f(2)=\frac14$, we can apply this twice to get $f(1)=1$. Then we have: $$Q(x,n):\frac1{f(x+n)}=n^2+2nx+\frac1{f(x)}$$for $n\in\mathbb N$ by induction. Now: $Q(1,n)\Rightarrow f(n)=\frac1{n^2}$. $Q\left(\frac1n,n\right)\Rightarrow\frac1{f\left(n+\frac1n\right)}=n^2+2+\frac1{f\left(\frac1n\right)}$ $P\left(n,\frac1n\right)\Rightarrow n^2f\left(\frac1n\right)^2+(1-n^4)f\left(\frac1n\right)-n^2=0\Rightarrow f\left(\frac1n\right)=n^2$ Let $x=\frac pq$. $Q\left(\frac1q,p\right)\Rightarrow\frac1{f\left(p+\frac1q\right)}=p^2+\frac1{q^2}+2x$ $P\left(p,\frac1q\right)\Rightarrow f(x)=\frac{\frac{q^2}{p^2}+q^4}{p^2q^2+1}\Rightarrow\boxed{f(x)=\frac1{x^2}}$, which works.

jasperE3 wrote: Let $p,q,n$ be arbitrary natural numbers. $P(x,1)\Rightarrow f(x+1)=\frac{f(x)}{f(x)+2xf(x)+f(1)}$ Since $P(1,1)\Rightarrow f(2)=\frac14$, we can apply this twice to get $f(1)=1$. Then we have: $$Q(x,n):\frac1{f(x+n)}=n^2+2nx+\frac1{f(x)}$$for $n\in\mathbb N$ by induction. Now: $Q(1,n)\Rightarrow f(n)=\frac1{n^2}$. $Q\left(\frac1n,n\right)\Rightarrow\frac1{f\left(n+\frac1n\right)}=n^2+2+\frac1{f\left(\frac1n\right)}$ $P\left(n,\frac1n\right)\Rightarrow n^2f\left(\frac1n\right)^2+(1-n^4)f\left(\frac1n\right)-n^2=0\Rightarrow f\left(\frac1n\right)=n^2$ Let $x=\frac pq$. $Q\left(\frac1q,p\right)\Rightarrow\frac1{f\left(p+\frac1q\right)}=p^2+\frac1{q^2}+2x$ $P\left(p,\frac1q\right)\Rightarrow f(x)=\frac{\frac{q^2}{p^2}+q^4}{p^2q^2+1}\Rightarrow\boxed{f(x)=\frac1{x^2}}$, which works. FTFY

Find all functions $ f: \mathbb{Q}^{+} \mapsto \mathbb{Q}^{+}$ such that: \[ f(x) + f(y) + 2xy f(xy) = \frac {f(xy)}{f(x+y)}.\] Solved with The new ThinkThink. The only solution is $\boxed{f(x)=\frac{1}{x^2}}$. This can be verified to work. We now prove it is the only solution. Let $P(x,y)$ denote the given assertion. $P(x,1): f(x)+f(1) + 2xf(x) = \frac{f(x)}{f(x+1)}$. Setting $x=1$ here gives $4f(1) = \frac{f(1)}{f(2)}\implies f(2) =\frac{1}{4}$. $P(2,2): 2f(2) + 8f(4) = 1\implies f(4) = \frac{1}{16}$. $P(3,1): f(3) + f(1) + 6f(3) = \frac{f(3)}{f(4)} = 16f(3)\implies f(1) = 9f(3)$. $P(2,1): \frac{5}{4} + f(1) = \frac{1}{4f(3)}\implies \frac{5}{4} + 9f(3) = \frac{1}{4f(3)}$. Multiplying both sides by $4f(3)$ gives \[36f(3)^2 + 5f(3) - 1 =0\implies (9f(3)-1)(4f(3) + 1) =0,\]since $f(3)>0$, we have $f(3) =\frac{1}{9}$. Thus, $f(1) = 9f(3) = 1$. $P(x,1): f(x)+2xf(x) + 1 = \frac{f(x)}{f(x+1)}$. Claim: $f(n) = \frac{1}{n^2}$ for any positive integer $n$. Proof: We induct with base cases $n=1,2,3,4$. Suppose $f(x) = \frac{1}{x^2}$ for $x\in \{1,2,3,\ldots,n\}$. We have $\frac{f(n)}{f(n+1)} = \frac{1}{n^2 } + \frac{2}{n} + 1 = \frac{(n+1)^2}{n^2}$. So $\frac{1}{n^2f(n+1)} = \frac{(n+1)^2}{n^2}\implies f(n+1)=\frac{1}{(n+1)^2}$, so the induction is complete. $\square$ Claim: For any integer $a>0$ and rational number $x>0$, we have \[f(x+a) = \frac{f(x)}{2axf(x) + a^2f(x) + 1}\]Proof: We will induct on $a$. Base case: We will show that this is true for $a=1$, or $f(x+1) = \frac{f(x)}{2xf(x) + f(x) + 1}$. By $P(x,1)$, we have \[2xf(x) + f(x) + 1 = \frac{f(x)}{f(x+1)},\]so $\frac{f(x+1)}{f(x)} = \frac{1}{2xf(x) + f(x) + 1}$, which implies $f(x+1) = \frac{f(x)}{2xf(x) + f(x) + 1}$. Inductive step: Suppose for all $n\in \{1,2,3,\ldots, a\}$ satisfy $f(x+n) = \frac{f(x)}{2nxf(x) + n^2f(x) + 1}$. We will show that \[f(x+a+1) = \frac{f(x)}{2axf(x) + 2xf(x) +(a+1)^2 f(x) + 1}\] $P(x+a,1): f(x+a) + 2(x+a)f(x+a) + 1 = \frac{f(x+a)}{f(x+a+1)}$, so \[f(x+a+1) = \frac{f(x+a)}{f(x+a) + 2(x+a)f(x+a) + 1}\] We have $f(x+a) = \frac{f(x) }{2axf(x) + a^2f(x) + 1}$, and \begin{align*} f(x+a) + 2(x+a)f(x+a) + 1 \\ = \frac{(2x+2a+1)f(x) + 2axf(x) + a^2f(x) + 1}{2axf(x) + a^2f(x) + 1} \\ =\frac{2xf(x) + 2af(x) + f(x) + 2axf(x) + a^2 f(x) + 1}{2axf(x) + a^2f(x) + 1} \\ \end{align*} So \[f(x+a+1) = \frac{f(x)}{2xf(x) + 2af(x) + f(x) + 2axf(x) + a^2f(x) + 1}= \frac{f(x)}{2axf(x) + 2xf(x)(a+1)^2 f(x) + 1}, \]as desired. $\square$ Let $Q(x,a)$ denote the assertion that \[f(x+a) = \frac{f(x)}{2axf(x) + a^2f(x) + 1},\]where $x\in \mathbb{Q}^{+}$ and $a\in \mathbb{N}$. $Q\left(\frac{1}{a},a\right): f\left(a+\frac{1}{a}\right) = \frac{f\left(\frac{1}{a}\right)}{2f\left(\frac{1}{a}\right) + a^2 f\left(\frac{1}{a}\right) + 1}$. This gives $\frac{1}{f\left(a+\frac{1}{a}\right)} = a^2 + 2 + \frac{1}{f\left(\frac{1}{a}\right)}$. If $a$ is an integer, \[P\left(a,\frac{1}{a} \right): \frac{1}{a^2} + f\left(\frac{1}{a}\right) + 2 = \frac{1}{f\left(a+\frac{1}{a}\right)}.\] Claim: If $x,y>0$ satisfy $x-y = \frac{1}{x} - \frac{1}{y}$, then $x=y$. Proof: We have \[x-y = \frac{1}{x} - \frac{1}{y} = \frac{y-x}{xy}\]If $x\ne y$, then dividing both sides by $x-y$ gives $\frac{-1}{xy} = 1$, not possible as $x,y>0$. So $x=y$. $\square$ Then \begin{align*} \frac{1}{a^2} + f\left(\frac{1}{a}\right) + 2 = a^2 + 2 + \frac{1}{f\left(\frac{1}{a}\right)} \\ \implies \frac{1}{a^2} + f\left(\frac{1}{a}\right) = a^2 + \frac{1}{f\left(\frac{1}{a}\right)} \\ \implies a^2 - f\left(\frac{1}{a}\right) = \frac{1}{a^2} - \frac{1}{f\left(\frac{1}{a}\right)} \\ \implies f\left(\frac{1}{a}\right) = a^2 \\ \end{align*} So $f\left(\frac{1}{a}\right) = a^2$ for all positive integers $a$. Let $p$ and $q$ be any positive integers. Using $f\left(\frac{1}{q}\right) = q^2$. \[Q\left(\frac{1}{q},p\right): f\left(p+\frac{1}{q}\right) = \frac{q^2}{2pq + p^2q^2 + 1}\] \[P\left(p,\frac{1}{q}\right): f(p) + q^2 + 2\cdot \frac{p}{q} f\left(\frac{p}{q}\right) = \frac{f\left(\frac{p}{q}\right)}{f\left(p+\frac{1}{q}\right)}\] So \[f(p) + q^2 + 2\cdot \frac{p}{q} f\left(\frac{p}{q}\right) = \frac{(2pq+p^2q^2 + 1)f\left(\frac{p}{q}\right)}{q^2}\] Multiplying both sides by $q^2$ gives \[q^2 f(p) + q^4 + 2pq f\left(\frac{p}{q}\right) = 2pq f\left(\frac{p}{q}\right) + p^2 q^2 f\left(\frac{p}{q}\right) + f\left(\frac{p}{q}\right)\] Simplifying and using $f(p) = \frac{1}{p^2}$ gives \[q^4 + \frac{q^2}{p^2} = (p^2q^2 + 1) f\left(\frac{p}{q}\right)\] Therefore, \begin{align*} f\left(\frac{p}{q}\right) \\ = \frac{q^4 + \frac{q^2}{p^2}}{p^2q^2 + 1} \\ = \frac{q^4p^2 + q^2}{p^4 q^2 + p^2} \\ = \frac{q^2(p^2q^2+1)}{p^2(p^2q^2+1)} \\ =\frac{q^2}{p^2} \\ =\frac{1}{\left(\frac{p}{q}\right)^2 } \\ \end{align*} Since $p$ and $q$ can be any positive integers, we have $\frac{p}{q}$ can be any positive rational numbers, so $f(x) = \frac{1}{x^2 }\forall x\in \mathbb{Q}^{+}$.