I don't understand what the 'uniquely' means. Anyway, Petra can squared each of the numbers and compare the differences of the results, with the given numbers to identify $xy$ and then $x+y$ and $x-y$ (and then $\frac x y$). Thus $x = \frac 1 2 (x+y + x-y)$ from which we easily get $y$ too. Pierre.

Pierre, I think something is wrong because if the first player took the number $-x$ and $-y$ you would get the same squares.....But not the same numbers.... Also, maybe I don't get something..... By the way, the name of the second player is Petya (the russian equivalent for Pierre ) , and not Petra , a woman's name

Xixas wrote: Vanya has chosen two positive numbers, x and y. DusT wrote: By the way, the name of the second player is Petya (the russian equivalent for Pierre ) , and not Petra , a woman's name

So? Am I right? Pierre.

I think you are right.... Strange that at the actual competition I didn't even notice the condition about the positiveness of the numbers I didn't solve the problem anyway (too many cases I think...)

pbornsztein wrote: I don't understand what the 'uniquely' means. Anyway, Petra can squared each of the numbers and compare the differences of the results, with the given numbers to identify $xy$ and then $x+y$ and $x-y$ (and then $\frac x y$). Thus $x = \frac 1 2 (x+y + x-y)$ from which we easily get $y$ too. How do you think: are there some cases? For example, it is possible $(x+y)^2-(xy)^2=2(x-y)^2$ Am I missing something?

Uniquely means "with no other possibilities", i.e. Petya can found numbers x and y, that had been chosen by Vanya homologously (don't know, if this word is proper here).

Xixas! I think, your translation is fine.