Birthday Functional Equations problems

Heyoo!!! Birthday FEs!!!!!! $11$ FEs!! Also I would be posting solutions to RG's FE handout, I am done with 10 prs :P!!

Problem: Find all functions $f :\Bbb R \rightarrow \Bbb R$ such that

$$2f (x) - 5f (y) = 8, \forall x, y \in \Bbb R$$

Solution: $$2f(x)-5f(y)=8$$

$$\implies 2f(x)-5f(x)=8$$

$$\implies f(x)=\frac{-8}{3}, \text{ a constant function }$$

We did this in Rohan Bhaiya's FE class..Oh btw the EGMO camp is sooo niceee! I am loving it!! It's such a big deal to be able to train and attend the camp with EGMO team members!

Problem: Find all functions $f :\Bbb R \rightarrow \Bbb R$ such that

$$f (x) + xf (1 -x) = x, \forall x\in \Bbb R.$$

Solution: $$f(x)+xf(1-x)=x$$

$$f(1-x)+(1-x)f(x)=1-x$$

This is actually in the linear equations in two variable form! $$x+ay=a$$

$$y+bx=b$$

Anyways,  $$f(x)+xf(1-x)=x$$

$$xf(1-x)+f(x)(1-x)x=(1-x)x$$

$$ \implies f(x)(x-x^2)-f(x)=-x^2\implies f(x)=\frac{-x^2}{x-x^2-1}=\frac{x^2}{x^2-x+1}$$

But verifying, this doesn't work.

Problem: Find all functions $f :\Bbb R \rightarrow \Bbb R$ such that

$$f (x - y) = f (x) + f (y) -2xy, \forall x,y \in \Bbb R$$

Solution: Let $P(x,y)$ be the assertion.

Then $$P(x,0)=f(x)=f(x)+f(0)\implies f(0)=0$$

$$P(x,x)=0=f(0)=2(f(x)-x^2)\implies \boxed{f(x)=x^2}$$

Which satisfies the solution..

Problem: Find all functions $f :\Bbb R \rightarrow \Bbb R$ for all $x,y$ such that

$$|x|f(y)+yf(x)=f(xy)+f(x^2)+f(f(y))$$

Solution: Let $P(x,y)$ be the assertion.

$$P(0,y)\implies yf(0)=2f(0)+f(f(y))\implies f(f(y))=f(0)(y-2)$$

Case1: $f(0)\ne 0$

Then, we get $f$ bijective. And $f(f(2))=0.$

But $P(1,1)\implies f(f(1))=0.$

Not possible byt injectivity.

Case2: $f(0)=0$

Then $P(0,y)\implies f(f(y))=0.$

Also, $P(x,0)\implies f(x^2)=0$ So $f(x)=0$ for all positive $x.$

Let $x$ be negative, $y$ positive.

$$P(x,y)\implies yf(x)=f(xy).$$

Let $x,y$ be negative..

$$P(x,y)\implies yf(x)=xf(y)\implies f(x)=-xf(1)=-xc.$$

Verifying, we get for $x$ negative and $y$ positive, $yf(x)=-xyf(1)=f(xy).$

And we get for $y$ negative and $x$ positive, $xf(y)=-xyf(1)=f(xy).$

Problem: Find all functions $\Bbb Z \rightarrow \Bbb Z$ that satisfy $f(0)=1$ and$$ f(f(n))=f(f(n+2)+2)=n$$

Solution: Since$$f(f(n))=n\implies f \text{ is bijective }$$Then easily, we get that$$f(n+2)+2=f(n)\implies f(n)=1-n \text{ for all even } n$$$$ \text{ for all even } n, f(f(n))=f(1-n)=n\implies f(1-n)=1-(1-n)$$

So $\boxed{ f(n)=1-n}.$

Problem: Find all functions $\Bbb Z \rightarrow \Bbb Z$ such that

$$f(x+f(y))=f(x)+y$$for all $x,y\in \Bbb Z.$

Proof: Let $P(x,y)$ be the assertion.

$$P(-f(y),0)\implies f(0)=f(-f(y))+y$$$$P(0,y)\implies f(f(y))=f(0)+y\implies f \text{ is bijective }$$$$f(x+f(0))=f(x)\implies x+f(0)=x\implies f(0)=0$$$$P(x,f(y))\implies f(x+y)=f(x)+f(y)\implies f(x)=f(x)\cdot f(1)$$Using the fact that $f(f(y))=y\implies f(1)^2=1\implies f(x)=\pm x.$

By pointwise trap, if $f(a)=a, f(b)=-b, a,b\ne 0\implies f(a-b)=a+b.$ Not possible.

So $\boxed{ f(x)=x \forall x, f(x)=-x\forall x}$

Problem[ IMO 2004/2]: Find all polynomials $f$ with real coefficients such that for all reals $a,b,c$ such that $ab+bc+ca = 0$ we have the following relations

Proof: Thanks to atul for help!$$ f(a-b) + f(b-c) + f(c-a) = 2f(a+b+c).$$Note that$$P(0,0,0)\implies  f(0) = 0$$And$$P(a,0,0) : f(a) = f(-a) \implies f \text{ is even.}$$

So$$f(x) = a_{2n}x^{2n} + a_{2n-2}x^{2n-2} + \ldots + a_6x^6 + a_4x^4 + a_2x^2+a_0, \text{ but } f(0)=0\implies a_0=0.$$

$$ f(a-b) + f(b-c) + f(c-a) = 2f(a+b+c)\implies (a-b)^{2n} + (b-c)^{2n} + (c-a)^{2n} = 2(a+b+c)^{2n}$$$$P(6x,3x,-2x)\implies 3^{2n}+5^{2n}+8^{2n}=2\cdot 7^{2n}\implies n\le 6 \implies n=1,2,3.$$

If $n=3\implies 3\equiv 0\pmod 7.$ Not possible. So $n=4, 2.$

So $\boxed{ ax^4+bx^2}$ is the solution. I haven't verified if they satisfy or not, but they should.

USAJMO 2015/P4: Find all functions $f:\mathbb{Q}\rightarrow\mathbb{Q}$ such that\[f(x)+f(t)=f(y)+f(z)\]for all rational numbers $x<y<z<t$ that form an arithmetic progression. ($\mathbb{Q}$ is the set of all rational numbers.)

Proof: $$f(a)+f(a+3d)=f(a+d)+f(a+2d)$$$$f(a-d)+f(a+2d)=f(a)+f(a+d)$$$$f(a-d)+f(a+3d)=2f(a+d)$$This is Jensen's Functional Inequality.

So $\boxed{f(x)=f(0)=xf(1)}$

Problem: Solve from $\Bbb Z_{>0} \rightarrow \Bbb Z_{>0}$ such that$$f(m+n)=f(m)+f(n)+mn$$

Solution:$$f(1+1)=f(1)+f(1)+1$$$$f(n-1+1)=f(n-1)+f(1)+n-1=f(n-2)+f(1)+f(1)+n-1+n-2=\dots= \boxed{nf(1)+\binom{n}{2}}$$

Problem: Find, all functions $f:\mathbb{R}\to\mathbb{R}$ satisfying

$$f(x^2+y)=f(f(x)-y)+4f(x)y$$for all real numbers $x$ and $y$.

Solution: Let $P(x,y)$ denote the assertion.

Using the hint: “DURR WE WANT STUFF TO CANCEL”

$$P\left(x,\frac{f(x)-x^2}{2}\right)\implies f(x)(f(x)-x^2)=0.$$So $f(x)=0$ or $f(x)=x^2$.

If all are mapped to zero or all are mapped to $x^2.$

Suppose not, then let $f(x)=x^2, f(x^2+y)=0, y\ne -x^2, x\ne 0.$

We have$$f(x^2-y)+4x^2y=0$$

So $f(x^2-y)$ is nonnegative.. Then$$f(x^2-y)=(x^2-y)^2\implies x^4+2x^2y+y^2=0\implies (x^2+y)^2=0\implies y=-x^2.$$Contradiction.

So $\boxed{f(x) = 0  \forall x,f(x)=x^2 \forall x}$

Problem[ IMO 2008 P4]: Find all functions $ f: (0, \infty) \mapsto (0, \infty)$ (so $ f$ is a function from the positive real numbers) such that

$$\frac {\left( f(w) \right)^2 + \left( f(x) \right)^2}{f(y^2) + f(z^2) } = \frac {w^2 + x^2}{y^2 + z^2}$$ for all positive real numbers $ w,x,y,z,$ satisfying $ wx = yz.$

Solution: By substituting $P(1,1,1,1)$ we get $f(1)=1$

by $P(x,1,x,1)$ we get $f(x^2)=f(x)^2$

Now considering $P(a^2,1,a,a)$ we get that $f(a)=a$ or $1/a$

For point wise trap : suppose there are reals $a,b$ satisfying $f(a)=a$ and $f(b)=1/b$ . By considering $P(a,b,ab,1)$ we get a contradiction

Hence $\boxed {f(x)=x}$ for all $x$

Enjoi!!

Sunaina💜