This problem is a troll. For any prime $p$ such that $p+2$ is also prime, at least one of $p, -p-2$ must be in any legendary set. This is easy to see since $xy-y^2-y+x$ can be factored as $(y+1)(x-y)$, so the previous claim can be verified by simply checking $4$ cases. As there are well over $8$ pairs of twin primes, the problem is easy now. $\square$

Pathological wrote: This problem is a troll. For any prime $p$ such that $p+2$ is also prime, at least one of $p, -p-2$ must be in any legendary set. This is easy to see since $xy-y^2-y+x$ can be factored as $(y+1)(x-y)$, so the previous claim can be verified by simply checking $4$ cases. As there are well over $8$ pairs of twin primes, the problem is easy now. $\square$ Please explain how are you arriving from $(x-y)(y+1)$ that all twin prime pairs must be present. Please tell.

Pathological wrote: This problem is a troll. Most of the students in israel don't have any olympiad-math knowledge The problems are designed to be solved in very very elementary and stupid way. You can call it "troll" but this is just because you were expecting something sophisticated...

Ignored.

yamamaya wrote: Consider $\{0, -2\}$, which should be legendary, but only has 2 numbers. You can't reach 1 (or any odd integer) from it

shalomrav wrote: yamamaya wrote: Consider $\{0, -2\}$, which should be legendary, but only has 2 numbers. You can't reach 1 (or any odd integer) from it OK, seems I misunderstood the question. Ignore my answer then.

I am not sure whether x and y should be different. If x and y are necessarily different, then only one element in a legendary set is available?? and {0,-2} is legendary, too ?? But it's no big deal, I would like to discuss the case "x and y should be different" first: Let the set be S $xy-y^2-y+x=(y+1)(x-y)$ I would like to prove that S=$\{0,-2\}$ is the only finite legendary set except for |S|=1. Now assume that S is finite. So there exist the maximum number and the minimum number of S, let them be M and m. (1)If |S|>2, then $M-m\ge2 \Rightarrow |(m+1)(M-m)|\ge2|m+1|\ge|m|$(equality holds when $M=0,m=-2$) or $m=-1$. Also, $|(M+1)(m-M)|=|(M+1)(M-m)|\ge2|M+1|\ge|M|$(equality holds when $M=-2,m=-4$) or $M=-1$. Obviously, It is impossible that $max(|(m+1)(M-m)|,|(M+1)(M-m)|)>max(M,m)$. So "$m=-1$" or "$M=-1$" or "$M=0,m=-2$" or "$M=-2,m=-4$". (i)If $-1\in S$ then $(-1+1)(x+1)=0\in S$. As a result $M\ge0$, then $m=-1$. $(0+1)(1-0)=1\in S \Rightarrow (1+1)(-1-1)=-4\in S$. But $-4<m=-1$, a contradiction. (ii)If $M=-2,m=-4$, then $(-4+1)(-2+4)=-6\in S$. Still a contradiction. (iii)$M=0,m=-2$ is legendary??? (2)If |S|=2. We only need to discuss the case "$M=m+1$".(If $M-m\ge2$, it also makes contradiction just like (1)). Now $(m+1+1)(m-(m+1))=-(m+2)\in S\Rightarrow m=1,M=2\Rightarrow (2+1)(1-2)=-3\in S$. Still a contradiction. (3)|S|=1 is legendary??? Therefore there are infinite many elements in S or $S=\{0,-2\}$ or |S|=1 Q.E.D. If x and y are not necessarily different, then it's true that there are infinite many elements in S