littletush wrote: determine all real-valued functions $f$ on $R$ satisfying $2f(x)=f(x+y)+f(x+2y)$ for all real numbers $x$ and $y\ge 0$. Let $P(x,y)$ be the assertion $2f(x)=f(x+y)+f(x+2y)$ (a) : $P(2x,x)$ $\implies$ $2f(2x)=f(3x)+f(4x)$ (b) : $P(x,x)$ $\implies$ $2f(x)=f(2x)+f(3x)$ (c) : $P(0,x)$ $\implies$ $2f(0)=f(x)+f(2x)$ (d) : $P(0,2x)$ $\implies$ $2f(0)=f(2x)+f(4x)$ (a)-(b)+4(c)-(d) : $f(x)=f(0)$ And so $\boxed{f(x)=c}$ $\forall x\ge 0$ and $f(x)=$ any value $\forall x<0$ which indeed is a solution

I guess because $y \ge 0$ is written in one LaTeX and not two (as in $y$ $\ge 0$), littletush intended it to be for all $x$ real and for all $y \ge 0$. Hence $f(x) = c$ for all $x$, not only for non-negative ones. [mod: the problem statement has now been edited to be clearer]

You can see here link: http://mathproblems123.wordpress.com/2011/04/20/romanian-tst-2011-problem-1/

here is my solution: Comparing $P(x,y)$ and $P(x,2y)$ we get $f(x+y)+f(x+2y)=f(x+2y)+f(x+4y)$ so $f(x+y)=f(x+4y)$. Therefore when $x=-y$ we get $f(0)=f(3y)$ so when $x>0$, we have $f(x)=f(0)=c$. If $a$ is a negative real then $P(a,-a)$ implies $2f(a)=f(0)+f(-a)=2c$ so it follows $f(x)=c$ is constant, for any constant $c$.

Here's another solution: $P(-t,t) \implies 2f(-t)=f(0)+f(t)$ $P(t,-t) \implies 2f(t)=f(0)+f(-t)$ Hence, $2f(-t)+f(0)+f(-t)=f(0)+f(t)+ 2f(t) \implies f(-t)=f(t)$ Putting this back into $P(-t,t)$ we get $f(t)=f(0)$ so $ \boxed{f(x)=c} $ which is indeed a solution

$f(0)=a$ ,so $2f(x)=f(-x+x)+f(2x-x)$ so $f(x)=f(0)=a$

shekast-istadegi wrote: $f(0)=a$ ,so $2f(x)=f(-x+x)+f(2x-x)$ so $f(x)=f(0)=a$ I dont understand : From $2f(x)=f(x+y)+f(x+2y)$, what are the values you used for $x,y$ in order to get your equality ?

prasanna1712 wrote: Here's another solution: $P(-t,t) \implies 2f(-t)=f(0)+f(t)$ $P(t,-t) \implies 2f(t)=f(0)+f(-t)$ Hence, $2f(-t)+f(0)+f(-t)=f(0)+f(t)+ 2f(t) \implies f(-t)=f(t)$ Putting this back into $P(-t,t)$ we get $f(t)=f(0)$ so $ \boxed{f(x)=c} $ which is indeed a solution In $P(x,y)$ one can only use $y\geq 0$, so both $P(-t,t)$ and $P(t,-t)$ cannot be simultaneously considered.

A quicker way to finish off my solution: After obtaining $f(x+y)=f(x+4y)$, compare the result of letting $x=-y$ and $x=-4y$. We obtain $f(3y)=f(0)=f(-3y)$ so $f(x)$ is clearly constant.

I think this is too easy for a TST even for a day $1$ number $1$.Plugg $x,y$ and then plugg $x,2y$ and obtain $f(x+y)=f(x+4y)$ and from this point it is obvious and it can be finished in milion ways,but still here is one:Plugg $x=-y$ to get $f(3x)=f(0)$ and $x=-4y$ to get $f(-3y)=f(0)$,so we have that $f(x)=c$ for any real $c$.

Let $g(x) = f(x)-f(0)$, then observe that by subtracting $2f(0)$ from both sides, $2g(x) = g(x+y) + g(x+2y)$, call this $P(x,y)$. Observe that $g(0)= 0$. Fix $y\ge 0$. Then, taking $P(-y,y)$ yields $2g(-y) = g(y)$. Then, take $P(0,y)$ to get $g(y)=-g(2y) = -2g(-2y)$. Now, take $P(-2y,y)$ to get $2g(-2y) = g(-y)$. But $2g(-2y) = -g(y)$, so $g(y)+g(-y) = 0$; since $g(y) = 2g(-y)$, this yields $g(y) = g(-y) = 0$ for all $y\ge 0$, hence $g(x) = 0$ for all $x$, so $f(x) - f(0) = 0$, or $f(x) = f(0)$, hence $f$ is constant, which works.