I'm going through solution for A5 from International Mathematical Olympiad 2015. Something's not right. Here's how it goes:
First, we assume all functions below map integers to integers.
Second, for any function $g$ and any nonzero integer $t$ we introduce an operator $$ \Delta_tg(x)=g(x+t)-g(x). $$
Third, we are given some very specific function $f$ and for all $x,a\in\mathbb{Z}$ we establish $$ \Delta_{f(x)}f(a)=\Delta_{f(x)}f(2x-a-f(x)). $$
Here's where it gets troublesome. They successively apply this last expression to every $a = b, b+f(x),\dots,b+(k-1)f(x)$, where $b,k$ are arbitrary positive integers, and then sum them up. Summing them up gives them $$ \Delta_{kf(x)}f(b)=\Delta_{kf(x)}f(2x-b-kf(x)). $$ Which I can't repeat. Instead I'm getting $$ \Delta_{kf(x)}f(b)=\Delta_{kf(x)}f(2x-b-f(x)). $$ Here's how I derive it for $k=2$. Left side: $$ \Delta_{f(x)}f(b)+\Delta_{f(x)}f(b+f(x))= $$ $$ =f(b+f(x))-f(b)+f(b+2f(x))-f(b+f(x))= $$ $$ =f(b+2f(x))-f(b)=\Delta_{2f(x)}f(b). $$ Right side: $$ \Delta_{f(x)}f(2x-b-f(x))+\Delta_{f(x)}f(2x-b-2f(x))= $$ $$ =f(2x-b-2f(x))-f(2x-b-f(x))+f(2x-b-3f(x))-f(2x-b-2f(x))= $$ $$ =f(2x-b-3f(x))-f(2x-b-f(x))=\Delta_{2f(x)}f(2x-b-f(x)). $$ I can't believe they made an error in official solutions, but I don't see any in mine either. Any suggestions? Thanks.
$\endgroup$ 51 Answer
$\begingroup$$$\begin{align}\left(\Delta_{kf(x)}f\right)(b)&=\sum_{j=0}^{k-1}\left(\Delta_{f(x)}f\right)(b+jf(x))\\ &=\sum_{j=0}^{k-1}\left(\Delta_{f(x)}f\right)(2x-b-jf(x)-f(x))\\ &=\sum_{j=0}^{k-1}\left(\Delta_{f(x)}f\right)\left(2x-b-kf(x)+(k-j-1)f(x)\right)\\ &=\sum_{i=0}^{k-1}\left(\Delta_{f(x)}f\right)\left(2x-b-kf(x)+if(x)\right)\\ &=\left(\Delta_{kf(x)}f\right)\left(2x-b-kf(x)\right) \end{align}$$
where $i=k-1-j$ in the substitution.
Your error is in calculating the right side values. The correct values are:
$$\begin{align}\left(\Delta_{f(x)}f\right)\left(2x-b-f(x)\right)&=f\left(2x-b\right)-f\left(2x-b-f(x)\right)\\ \left(\Delta_{f(x)}f\right)\left(2x-b-2f(x)\right)&=\left(2x-b-f(x)\right)-\left(2x-b-2f(x)\right) \end{align}$$
And you see the add up to $\left(\Delta_{f(x)}f\right)(2x-b-2f(x)).$
In your calculation, you're using that $\Delta_t f(x)=f(x-t)-f(x)$, which is both the wrong direction and the wrong order.
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