$ 3x+y=a $ $ x+2y=b $ $ Then $ $ 4x+3y=a+b $ $ f(a+b)=f(a)+f(b) $

Feridimo wrote: $ 3x+y=a $ $ x+2y=b $ $ Then $ $ 4x+3y=a+b $ $ f(a+b)=f(a)+f(b) $ For all integers "x,y"*

RnstTrjyn wrote: Feridimo wrote: $ 3x+y=a $ $ x+2y=b $ $ Then $ $ 4x+3y=a+b $ $ f(a+b)=f(a)+f(b) $ For all integers "x,y"* ...........

@above You know what, I suspect it! Once you choose a number say $203\in N$, you get a constrain $3x+y=203\implies y=203-3x\implies x+2y=406-5x$. So for $b$ you can't choose the numbers that are incongurent to $406$ in $\mathbb{Z}/5\mathbb{Z}$. Let me know if I am wrong

solving for x and y from a and b must give int solutions

RishiNandha_M wrote: solving for x and y from a and b must give int solutions Really? Try solving $$3x+y=2$$$$x+2y=3$$

na im saying u gotta make that a contraint when you solve the fe

I see I thought that you were in support of #4

$ y=-x $ $ f(x)=f(2x)+f(-x) $ $ and $ $ f(-x)=f(-2x)+f(x) $ $ Then $ $ f(-x)=-f(x) $ $ f(x)=f(2x)+f(-x)=f(2x)-f(x) $ $ f(2x)=2f(x) $ $ f(4x)=2f(2x)=4f(x) $ $ y=0 $ $ f(4x)=f(3x)+f(x)=4f(x) $ $ f(3x)=3f(x) $ $ Then $ $ f(nx)=nf(x) $ ;$ x=1 $; $ f(n)=nf(1)=an $

Feridimo wrote: $ y=-x $ $ f(x)=f(2x)+f(-x) $ $ and $ $ f(-x)=f(-2x)+f(x) $ $ Then $ $ f(-x)=-f(x) $ $ f(x)=f(2x)+f(-x)=f(2x)-f(x) $ $ f(2x)=2f(x) $ $ f(4x)=2f(2x)=4f(x) $ $ y=0 $ $ f(4x)=f(3x)+f(x)=4f(x) $ $ f(3x)=3f(x) $ $ Then $ $ f(nx)=nf(x) $ ;$ x=1 $; $ f(n)=nf(1)=an $

and how do you use induction? for instance $f(5x)=5f(x)$ how do you get this?

arshiya381 wrote: Feridimo wrote: $ y=-x $ $ f(x)=f(2x)+f(-x) $ $ and $ $ f(-x)=f(-2x)+f(x) $ $ Then $ $ f(-x)=-f(x) $ $ f(x)=f(2x)+f(-x)=f(2x)-f(x) $ $ f(2x)=2f(x) $ $ f(4x)=2f(2x)=4f(x) $ $ y=0 $ $ f(4x)=f(3x)+f(x)=4f(x) $ $ f(3x)=3f(x) $ $ Then $ $ f(nx)=nf(x) $ ;$ x=1 $; $ f(n)=nf(1)=an $

and how do you use induction? for instance $f(5x)=5f(x)$ how do you get this? We het the $f(x)=kx$, from the $ f(a+b)=f(a)+f(b) $.

TuZo wrote: arshiya381 wrote: Feridimo wrote: $ y=-x $ $ f(x)=f(2x)+f(-x) $ $ and $ $ f(-x)=f(-2x)+f(x) $ $ Then $ $ f(-x)=-f(x) $ $ f(x)=f(2x)+f(-x)=f(2x)-f(x) $ $ f(2x)=2f(x) $ $ f(4x)=2f(2x)=4f(x) $ $ y=0 $ $ f(4x)=f(3x)+f(x)=4f(x) $ $ f(3x)=3f(x) $ $ Then $ $ f(nx)=nf(x) $ ;$ x=1 $; $ f(n)=nf(1)=an $

and how do you use induction? for instance $f(5x)=5f(x)$ how do you get this? We het the $f(x)=kx$, from the $ f(a+b)=f(a)+f(b) $. but a and b are not just ANY number they are a specific set of numbers and not $Z$ lets just say NO prime in the form of $4k+1$ can be constructed in this function

nevermind

\begin{align*} P(0,0) &\Longrightarrow f(0) =0\\ P(0, x) &\Longrightarrow f(3x) = f(2x) + f(x)\\ P(x, -x) &\Longrightarrow f(x) = f(2x) + f(-x) \end{align*}Using the second identity in $P(x, -2x)$ we get $$f(-2x) = f(x) + f(-3x) = f(x) + f(-x) + f(-2x) \Longrightarrow f(-x) =-f(x)$$Then from the third one we get $f(2x) = 2f(x)$ which also gives us $f(4x)=2f(2x) = 4f(x)$. From the second one we also have $f(3x) = 3f(x)$. Now, let's use induction to prove that $$f(k) =\begin{cases} kf(1) &\text{ if } 5\nmid k\\ \frac{k}{5}f(5) &\text{ if } 5\mid k \end{cases}$$for all positive integers $k$. We can get cases $1\leq k\leq 5$ by taking $x=1$ above. So, suppose the claim holds up to some $k>5$. If $k=3m$ for some $m\geq 2$, then (a) if $m\nmid 5$, then $$f(k) = f(3m)= f(2m) + f(m) = 2mf(1) + mf(1) = 3mf(1) = kf(1)$$where we used induction hypothesis because $m<2m<3m=k$, $\,5\nmid m$ and $5\nmid 2m$. (b) if $m\mid 5$, then $$f(k) = f(3m) = f(2m) + f(m) = \frac{2m}{5}f(5) + \frac{m}{5}f(5) = \frac{3m}{5}f(5) = \frac{k}{5}f(5)$$where we used induction hypothesis because $m<2m<3m=k$, $\,5\mid m$ and $5\mid 2m$. If $k=3m+1$ for some $m\geq 2$, then (a) If $m\not\equiv 3 \pmod{5}$, then $$P(1, m-1)\Longrightarrow f(k) = f(3m+1) = f(m+2) + f(2m-1) = (m+2)f(1) + (2m-1)f(1) = (3m+1)f(1) = kf(x)$$where we used induction hypothesis because $m+2, 2m-1 < 3m+1=k$ , $\,5\nmid m+2$ and $5\nmid 2m-1$. (b) If $m\equiv 3 \pmod{5}$, then $$P(1, m-1)\Longrightarrow f(k) = f(3m+1) = f(m+2) + f(2m-1) = \frac{(m+2)}{5}f(1) + \frac{(2m-1)}{5}f(1) = \frac{(3m+1)}{5}f(1) = \frac{k}{5}f(x)$$where we used induction hypothesis because $m+2, 2m-1 < 3m+1=k$ , $\,5\mid m+2$ and $5\mid 2m-1$. If $k=3m+2$ for some $m\geq 2$, then (a) If $m\not \equiv 1\pmod{5}$, then $$P(-1, m+2)\Longrightarrow f(k) = f(3m+2) = f(m-1) + f(2m+3) = (m-1)f(1) + (2m+3)f(1) = (3m+2)f(1) = kf(1)$$where we used induction hypothesis because $m-1, 2m+3 < 3m+2=k$, $\,5\nmid m-1$ and $5\nmid 2m+3$. (b) If $m \equiv 1\pmod{5}$, then $$P(-1, m+2)\Longrightarrow f(k) = f(3m+2) = f(m-1) + f(2m+3) = \frac{(m-1)}{5}f(1) + \frac{(2m+3)}{5}f(1) = \frac{(3m+2)}{5}f(1) = \frac{k}{5}f(1)$$where we used induction hypothesis because $m-1, 2m+3 < 3m+2=k$, $\,5\mid m-1$ and $5\mid 2m+3$. Since $f(-x)=-f(x)$ and $f(0)=0$ we get $$f(x) =\begin{cases} xf(1) &\text{ if } 5\nmid x\\ \frac{x}{5}f(5) &\text{ if } 5\mid x \end{cases}$$for all $x\in \mathbb{Z}$. To verify first note that $$5\mid 4x+3y \Longleftrightarrow 5 \mid 8x+6y \Longleftrightarrow 5 \mid 3x+y \Longleftrightarrow 5 \mid 6x+2y \Longleftrightarrow 5 \mid x+2y$$So either $$f(4x+3y)= (4x+3y)f(1) = (3x+y)f(1) + (x+2y)f(1) = f(3x+y) + f(x+2y) $$or $$f(4x+3y)= \frac{(4x+3y)}{5}f(5) = \frac{(3x+y)}{5}f(5) + \frac{(x+2y)}{5}f(5) = f(3x+y) + f(x+2y) $$

For example a=x+2y can be any integer when x,y there are any integers, so not only specified numbers.

RnstTrjyn wrote: Let $Z$ be the set of all integers. Find all the function $f: Z->Z$ such that $f(4x+3y)=f(3x+y)+f(x+2y)$ For all integers $x,y$ Let $P(x,y)$ be the assertion $f(4x+3y)=f(3x+y)+f(x+2y)$ Let $a=f(1)$ and $c=f(5)$ $P(0,0)$ $\implies$ $f(0)=0$ $P(2x-1,3-x)$ $\implies$ $f(5(x+1))=f(5x)+c$ and so $f(5x)=cx$ Adding $P(0,x)$ with $P(-x,2x)$, we get $f(-x)=-f(x)$ and $f(x)$ is odd $P(x,-x)$ $\implies$ $f(2x)=2f(x)$ $P(0,x)$ $\implies$ $f(3x)=3f(x)$ $P(1,x-1)$ $\implies$ $f(3x+1)=f(x+2)+f(2x-1)$ $P(2,x-2)$ $\implies$ $f(3x+2)=f(x+4)+f(2x-2)$ Note that these 3 equations show that knowledge of $f(x)$ over $\{0,1,2,3,4,5\}$ gives full knowledge over $\mathbb N$, and so, since odd, over $\mathbb Z$ Note also that in each equations, either neither argument is $\equiv 0\pmod 5$, either all are. About knowledge of $f(x)$ over $\{0,1,2,3,4,5\}$ : $f(0)=0$ $f(1)=a$ $f(2)=2f(1)=2a$ $f(3)=3f(1)=3a$ $f(4)=2f(2)=4a$ $f(5)=c$ From there, easy induction gives $\boxed{f(5x)=cx\text{ and }f(x)=ax\quad\forall x\not\equiv 0\pmod 5}$ Which indeed is a solution, whatever are $a,c\in\mathbb Z$

Excuse me , Why we can't use cauchy at start ?

Hedy wrote: Excuse me , Why we can't use cauchy at start ? Feridimo wrote: $ 3x+y=a $ $ x+2y=b $ $ Then $ $ 4x+3y=a+b $ $ f(a+b)=f(a)+f(b) $ Jupiter_is_BIG wrote: @above You know what, I suspect it! Once you choose a number say $203\in N$, you get a constrain $3x+y=203\implies y=203-3x\implies x+2y=406-5x$. So for $b$ you can't choose the numbers that are incongurent to $406$ in $\mathbb{Z}/5\mathbb{Z}$. Let me know if I am wrong $f: Z->Z$ $ x,y-$integer....

Feridimo wrote: Hedy wrote: Excuse me , Why we can't use cauchy at start ? Feridimo wrote: $ 3x+y=a $ $ x+2y=b $ $ Then $ $ 4x+3y=a+b $ $ f(a+b)=f(a)+f(b) $ Jupiter_is_BIG wrote: @above You know what, I suspect it! Once you choose a number say $203\in N$, you get a constrain $3x+y=203\implies y=203-3x\implies x+2y=406-5x$. So for $b$ you can't choose the numbers that are incongurent to $406$ in $\mathbb{Z}/5\mathbb{Z}$. Let me know if I am wrong $f: Z->Z$ $ x,y-$integer.... If we change to $f : \mathbb{Q} \rightarrow \mathbb{Q}$ is will be finish ???

Hedy wrote: If we change to $f : \mathbb{Q} \rightarrow \mathbb{Q}$ is will be finish ??? Yes, because for arbitrary $a,b \in \mathbb{Q}$ we can choose $x=\frac{2a-b}{5}$ and $y=\frac{3b-a}{5}$, then $3x+y=a$ and $x+2y=b$. Thus we get $$f(a+b)=f(a)+f(b),$$for any $a,b \in \mathbb{Q}$. And the Cauchy equation on $\mathbb{Q}$ has only the solutions $f(x)=cx$ with $ c \in \mathbb{Q}$.

Gryphos wrote: Hedy wrote: If we change to $f : \mathbb{Q} \rightarrow \mathbb{Q}$ is will be finish ??? Yes, because for arbitrary $a,b \in \mathbb{Q}$ we can choose $x=\frac{2a-b}{5}$ and $y=\frac{3b-a}{5}$, then $3x+y=a$ and $x+2y=b$. Thus we get $$f(a+b)=f(a)+f(b),$$for any $a,b \in \mathbb{Q}$. And the Cauchy equation on $\mathbb{Q}$ has only the solutions $f(x)=cx$ with $ c \in \mathbb{Q}$. Try $ a=2,b=1 ., x=0.6 $ok?

Very straightforward for P5. All answers are $$f(x) = \begin{cases} c_1x & 5\nmid x \\ \frac{c_2x}{5} & 5\mid x\end{cases}$$where $c_1,c_2\in\mathbb{Z}$ are constants. To verify this, unravel the functional equation to $$f(a+b)=f(a)+f(b)\text{ whenever } a\equiv 3b\pmod 5$$which makes this solution obviously works. Now we prove that they are all solutions. We divide the proof into two steps. Step 1: The difference method. Let $g(x) = f(x+5)-f(x)$. Clearly $g$ is constant on $5\mathbb{Z}$ as we can plug in $5\mid a,b$. Now we claim that Claim: $g$ is constant on $K = \mathbb{Z}\setminus 5\mathbb{Z}$. Proof: Whenever $a\equiv 3b\pmod 5$, we have $$f(a)+f(b+5) = f(a+b+5) = f(a+5) + f(b)$$thus $g(a)=g(b)$. Using this for one more time gives $g(a)=g(b)=g(a+5)$ thus $g$ has period $5$. However, notice that $$g(1) \stackrel{\times 3}{=} g(3) \stackrel{\times 3}{=} g(4) \stackrel{\times 3}{=} g(2)$$Hence $g$ is constant on $K$ as desired. Step 2: The system of equations. Let $g(x)=k$ for every $x\in K$. Using the functional equation, we have \begin{align*} f(1)+f(3) &= f(4) \\ f(2)+f(1) &= f(3) \\ f(3)+f(4) &= f(2)+k \\ f(4)+f(2) &= f(1)+k \end{align*}Directly solving the system of equation (treating $k$ as a constant) gives $f(n)=\tfrac{nk}{5}$ for every $n\in\{1,2,3,4\}$. Step 1 extends this to all element in $K$.

It turns out to be different for numbers divisible by 5 and different for other numbers.

Feridimo wrote: $ y=-x $ $ f(x)=f(2x)+f(-x) $ $ and $ $ f(-x)=f(-2x)+f(x) $ $ Then $ $ f(-x)=-f(x) $ $ f(x)=f(2x)+f(-x)=f(2x)-f(x) $ $ f(2x)=2f(x) $ $ f(4x)=2f(2x)=4f(x) $ $ y=0 $ $ f(4x)=f(3x)+f(x)=4f(x) $ $ f(3x)=3f(x) $ $ Then $ $ f(nx)=nf(x) $ ;$ x=1 $; $ f(n)=nf(1)=an $

This can't be handled with wholely induction since the base case for $n=5$ doesn't work: $5$ can't be written in the form $4a+3b$, where $a,b$ are nonnegative integers. Nevertheless, we can prove that $f(5k)=kf(5)$, and $f(zx)=zf(x)$ for all $z$ that are not divisible by $5$ by induction. Thus the answer follows.