We claim it is doable if and only if: - at least one of $u$ and $v$ is positive - if exactly one is positive other is nonzero - $(u,v)$ is not $(1,-1)$ or $(-1, 1)$ From (i), we can write $u$, $v$ and $u+v$. Then, $ux^2+(u+v)x+v=0$ has $x=-1$ as a root. Hence, we can write $-1$. This allows us to write any number at most $v$. However, if $v$ is positive and unequal to $1$, then we can write $v+v+\dots+v-1-1-\dots 1$ to get any positive integer. This can be done: $v\rightarrow v-1 \rightarrow 2v-1 \rightarrow \dots \rightarrow kv-1\rightarrow \dots \rightarrow kv-l$. From all positive integers, we can easily get all negative integers using (ii). If $v=1$, $u<0$, then if $u\leq -2$, we can construct $-2$ and $-1$ and thus $x^2-x-2$ from which we obtain $2$. Then, proceed as before to finish. If $(u, v)=(1, -1)$, we can check that it doesn't work. If $u$ and $v$ are both nonpositive, from (i) we cannot get any positive integer, and from (ii) we still cannot get any positive integer. If $u$ is zero, then obviously we can't write any other number. @below thanks seems like i rushed it too much

LLL2019 wrote: We claim it is doable if and only if at least one of $u$ and $v$ is positive. From (i), we can write $u$, $v$ and $u+v$. Then, $ux^2+(u+v)x+v=0$ has $x=-1$ as a root. Hence, we can write $-1$. This allows us to write any number at most $v$. However, if $v$ is positive, then we can write $v+v+\dots+v-1-1-\dots 1$ to get any integer. If $u$ and $v$ are both nopositive, from (i) we cannot get any positive integer, and from (ii) we still cannot get any positive integer. I think we also used that $u+v \not = u,v$, so $u,v \not = 0$? P.S. also how we can get any integer using only sums $v+v+\dots+v-1-1-\dots 1$? We can sum only distinct numbers. For example if $u=-1,v=1$, then numbers on board will be only $1,0,-1$, I think...

And also u cant have u=1 and v=-1 since we only get -1,0,1 from (i) and same thing from (ii).

Tintarn wrote: Two different integers $u$ and $v$ are written on a board. We perform a sequence of steps. At each step we do one of the following two operations: (i) If $a$ and $b$ are different integers on the board, then we can write $a + b$ on the board, if it is not already there. (ii) If $a$, $b$ and $c$ are three different integers on the board, and if an integer $x$ satisfies $ax^2 +bx+c = 0$, then we can write $x$ on the board, if it is not already there. Determine all pairs of starting numbers $(u, v)$ from which any integer can eventually be written on the board after a finite sequence of steps. Neat solution. Solved with Mirhabib. İt is easy to observe that $0$ cannot be written initially since we won't be able to do any of the operations. Now, note that letting $(a,b,c)=(u,u+v,v)$ gives us that $x=-1$ is a solution. So $-1$ is written on the board after $2$ steps Look at the case that at least of $u,v$ is positive. İ will first look at when both are positive . Keeping the process $v+u,+u,+u....$ we get that there can be arbitrarily large integer written on the board. Now keep doing the process $(-1)$ to get that all integers greater than $-2$ are written on the board since after getting $-1$ we cannot continue this . However, as we can select $3$ positive integers such that $4a+c=2b$, we get that those 3 numbers are the coefficients of the second process with root $-2$. So, keeping the process $-1,-1,-1,...$ we get that all negative integers are also on the board. İf only one of u, v, say $u$ is positive and $u$ is bigger than 1, then we get that $u-1,2u-1,...$ are on the board and so we can finish like the previous case. İf $u=1,v<-1$, then it is easy to see that all the numbers not bigger than 1 can be written on the board. However, we can generate a quadratic equation and get $2$ as a root. This yields that we get all the positive numbers on the board, and so finish like the other cases. For the pair (u,v)=(1,-1) we see that it is impossible to get other numbers on the board except $-1,0,1$. This gives the contradiction. Now, finally, note that if both $u,v$ are negative integers, then both operations cannot lead to a positive integer. So all the integers cannot be written on the board. Done.

Let's assume, without loss of generality, that $u<v$. First, we note that if one of $u$ or $v$ is equal to $0$, then we cannot generate a third number other than them to use operation (ii). Also, if $v<0$, i.e. both of them are negative, then $u+v<0$, and we cannot generate a positive number, since the solutions of $ax^2+bx+c=0$ with $a,b,c<0$ cannot be positive. If $u=-1$, and $v=1$, then we can only get $u+v=0$, and we are stuck. For any other choice of $u$ or $v$, we can write any integer on the board. Indeed, first consider $u\ne -1$ or $v\ne 1$ and $u<0<v$. Indeed, if $u=-1$ and $1<v$, we can get all numbers $v-1$, $v-2$, ...,$1$, $0$ using operation (i). Having generated $1$, we can use it to generate any positive integer. Since $-v$ is a solution of $x^2+(v+1)x+v=0$, we can generate all negative integers from $-v$ to $-2$ using operation one with $-v$ and $1$, and all integers less than $-v$ using operation (i) with $-v$ and $-1$. If $u<-1$ and $v=1$, then we can generate $u+v$ different than $u,v$, and which will give us $-1$ as a solution of $ux^2+(u+v)x+v=0$. Then we can generate all integers smaller than $u$ using operation (i) with $-1$, and all missing numbers between $u$ and $1$ using operation (i). Since $2$ is an integer solution of $x^2-x-2=0$, we can get $2$ and hence any integer, as in the previous case. If $u<-1$ and $1<v$, then we can use operation (ii) to obtain $-1$, and we are done from the $u=-1$ and $1<v$ case. If $u>0$, then $v\geq 2$, and we can generate $-1$ from $ux^2+(u+v)x+v=0$. This will allow us to use operation (i) to generate all positive integers and 0. Since $-c$ is a solution of $x^2+(c+1)x+c=0$, we can also generate any integer smaller than or equal to $-2$. The desired starting pairs $(u,v)$ are the ones with $\min\{u,v\}>0$ or $\min\{u,v\}<0<\max\{u,v\}$ except for the pair $(-1,1)$.

Clearly $u$ and $v$ are nonzero. First, using (i) only, create the biggest list of numbers possible. It is not really important what the values are, only that: clearly there exist three distinct $a$, $b$, and $a+b$ on the board. Then since \[a(-1)^2+(a+b)(-1)+b=0\]it follows that $-1$ is on the board. If one of $u$ and $v$ is positive, we are already done: we can get arbitrarily positive values and combine with $-1$ to get any integer. If $u$ and $v$ are both negative, the set of numbers on the board will always be completely negative: The first operation will not create a nonnegative integer. The second operation will not either, as $c$ is negative and $ax^2$ and $bx$ are nonpositive. The solution set is all $u$ and $v$ nonzero with at least one positive integer. okay i'm back for corrections L me There are special cases when at least one of $u$ and $v$ is positive. Specifically my "clearly" statement in the second section was wrong. It is true that $-1$ is in the set; however if there is one positive value and it is equal to $1$ we must dive a little deeper. If the negative value is not $-1$ then we can still get sufficiently negative numbers. Then we can produce a positive $k$ if we can find negative $x$ and $y$ such that \[1(k^2)+x(k)+y=0\implies kx+y=-k^2.\]Hence $x=-(k-1)$ and $y=-k$ will work just fine. Then we can make all sufficiently positive numbers and thus all integers. Now we have the last case: if $(u,v)$ is $(-1,1)$ or $(1,-1)$ we fail as we can only produce the set $\{-1,0,1\}$ and nothing else. $\blacksquare$

My solution is the same as #2. If you just play around with small cases you will be able to generalize the solution

I claim that the answer is: None of the $u,v$ is zero, at least one of $u,v$ is positive, $u,v \neq (1,-1)$ If one of $u,v$ is zero, then we can not make the first step If none of $u,v$ if positive, we never add a non-negative value If $u,v=(1,-1)$, then we can add only $0$. Now suppose that $u,v$ is not of one of three forms above. Then add $u+v$. Because $x=-1$ is a solution of the equation $ux^2+(u+v)x+v=0$, now we have $-1$ on the board. Subtract from positive number on the board $-1$ some number of times such that we get $1$. Now we have $-1,1$ on the board and also some other number(one of $u,v$). We can then add either all numbers $\geq -1$ or $\leq 1$. In first case $x=-2$ is the solution of $x^2+2x+0=0$, in the second $x=2$ is the solution of $x^2-2x+0$, thus we can add all integer numbers.

Let a pair $(u,v)$ which allows us to write the entire set of integers on the blackboard good. We claim that all pairs $(u,v)$ such that $u,v\neq 0$ , atleast one of $u,v$ is positive and $(u,v) \neq (-1,1),(1,-1)$ are good. First we eliminate the bad pairs. Clearly, if $u=0$ or $v=0$, then no other new numbers can be created using move (i) and thus, we cannot create any other new numbers. Further, if the starting pair is $(1,-1)$, the only new number we can create using move (i) is 0. Thus, the only polynomials of the form $ax^2+bx+c$ we can form are $x^2-1$ (which has roots $1,-1$), $-x^2+1$ (which has the same roots), $-x^2+x$ (which has the roots $0$ and $-1$), $x^2-x$ (which has the same roots), $x-1$ (which has the root $1$) and $-x+1$ (which has the same root). Thus, no new numbers can be created as claimed. Further, if both $(u,v)$ are negative, any new numbers formed by rule (i) will also be negative. Further, any polynomial of the given form will have negative coefficients and thus, \[-ax^2-bx-c=0 \implies - (ax^2+bx+c)=0\]for positive $a,b,c$. Thus, if this factors as $-(x-r_1)(x-r_2)$ we see that $r_1$ and $r_2$ must both be negative for $a,b,c$ to all be positive. Thus, the roots of this equation are negative which implies that rule (ii) also creates only negative numbers. Now, we show that all other numbers are good. We first prove the following claim. Claim : All pairs of the form $(1,n)$ for some integer $n$ are good. Proof : Note that using rule (i), we can write all integers greater than $n$. In particular, we can create arbitrarily large positive integers. Now, this means for all sufficiently large positive $k$, we can write the numbers $2k$ and $k^2$ on the board. Thus, \[x^2+2kx+k^2=0 \implies (x+k)^2=0\]which allows us to write $-k$ on the board as well for sufficiently large positive $k$. Then, using rule (i), we can create all integers greater than $-k$ as well which implies that we can write all integers on the board using a finite number of steps. Now, consider a initial pair $(u,v)$ where $u>0$. Then, we can write $k=u+v$ on the board. Further, we can also clearly write $ul+v$ for all positive integers $l$, so consider sufficiently large $l$ such that $r=ul+v>1$. Now, we can also write $k+r$ as it is a linear combination of $u$ and $v$ with positive integral coefficients. Then, since we can write $k$ , $r$ and $k+r$, considering the polynomial, \[kx^2+(k+r)x+r = 0 \implies (kx+r)(x+1) =0\]which allows us to write $-1$ on the board. But now, applying rule (i) on $u$ and $-1$ we obtain $u-1$, continuing this likewise, we can obtain $1$. Once $1$ is written on the board, by the previous claim, we know that we can write all integers on the board which finishes the proof.

is this A or C? We claim that the only cases that will sao-ji are 1. both $u,v$ are negative 2. one of $u,v$ is zero 3. $\{u,v\}=\{1,-1\}$ Let's prove that these cases will sao-ji 1. if $a,b$ are negative, then $a+b$ is negative. If $a,b,c$ are negative, then $ax^2+bx+c$ only has negative roots. So we can only write negative numbers. 2. We can't write any new numbers in this case 3. We're stuck with the set $\{-1,0,1\}$ Now let's prove that everything else works. If $u,v$ are both positive, then we can first create $u+v$, and note that $ux^2+(u+v)x+v=(ux+v)(x+1)$, therefore we can create $-1$ and all negative numbers. We can also create $1$ and all positive numbers. If $u$ is negative and $v$ is positive, then there exists $\ell$ such that $u+\ell v$ is positive. Note that $(u+\ell v)x^2+(u+(\ell+1)v)x+v=((u+\ell v)x+v)(x+1)$ and we can create $-1$ and $1$ just like the last case.

Replace $(u,v)$ with $(a,b)$. Answer: all $a,b$ different such that $\max(a,b)>0$ and $a,b$ different from $0$ and $(a,b)$ different from $(1,-1)$. Why $a,b$ must satisfy these conditions: if $b=0$ then the only numbers we have on the board are $a$ and $0$( because $a+b=a$ and we cannot select $3$ different numbers). Now if by contradiction $a,b<0$ then by performing 1) we will have only negative numbers and by performing $2)$ if we have $x\ge0$ root of $ax^2+bx+c=0$ then $(-a)x^2+(-b)x+(-c)=0$ and all terms are $>0$,contradiction. Why these work:we take 2 cases: Case 1:$a+b$ different from $0$ We claim by induction on $n\in \mathbb N^*$ that $n(a+b)$ is on the board. Base case: $n=1$: just do $1)$ for $a$ and $b$ Induction step: Suppose we have $n(a+b)$ on the board.We have $2$ cases : Case 1: $n(a+b)$ is different form both $a$ and $b$. Perform $n(a+b)$ with $a$ to get $(n+1)a+nb$. We should have $(n+1)a+nb=b$ otherwise we are done. Similary $(n+1)b+na=a$.Adding we get $a+b=0$ and $a=b$ false.So we are done in this case. Case 2: $n(a+b)=b$ then perform $n(a+b)$ with $a$ to get $n(a+b)+a$. If it is different from $b$ we are done. Else $n(a+b)+a=b$ so $a=0$ false. So the induction is proved. Now fix any $x<0$ and choose $n,m,l\in \mathbb N^*$ such that $nx^2+mx+l=0$ and $n,m,l$ are different(for example $n=1$, $m$ very big such that $x^2+mx<0$ and $l$ the difference which is $>0$). Then just choose $n(a+b),m(a+b),l(a+b)$ so we can choose $x$. In conclusion all negative numbers are in the set.Then just choose $max(a,b)$ and $1-max(a,b)$ which is negative to get $1$ is in the set, then choose $(1,-1,-2)$ to get $2$ is in the set and then just use the number $1$ to get $3,4,5\dots$ are in the set,done !. Case 2: $a+b=0$; $(a,b)$ different from $(1,-1)$. If we get $2$ numbers $x$ and $y$ such that $x+y$ differnt from $0$ doing the same as above we are done. we have $(a,b,0)$ on the board and looking at the root of $ax^2+bx+0=0$ we get $1$ is on the board. Then $a+1$ or $b+1$ is different from $0$ and we are done. Comment: very technical problem for $P_1$. The idea was easy but the write-up gave me headaches

Solved with brainfertilzer. The answer is all $u,v$ such that $uv \not \in \{-1,0\}$ (essentially they are both nonzero and can't be $-1,1$ in some order) and at least one of $u,v$ is positive. First we show that nothing else works. If $uv = 0$, then we clearly can't get a third element from (i), meaning we can't apply (ii). Hence $uv \ne 0$. Now if $uv = -1$, then WLOG that $u = -1$ and $v = 1$. We get from operation (i) that $0$ can be on the board, and operation (i) alone doesn't give any more elements. Now, suppose we could operate (ii) and get another number. WLOG that $a \ne -1$ (since we can replace $a,b,c$ with $-a,-b,-c$). If $a =0 $, then $bx + c = 0$, so $x$ must be $-1$ or $1$, contradiction. If $a = 1$, we either have $x^2 - 1 = 0$ or $x^2 - x = 0$, both of which imply $x\in \{-1,0,1\}$, contradiction. Thus, $uv \ne -1$. Now if $u,v$ were both negative, then notice that the first operation keeps everything negative and there can't be a positive root of $ax^2 + bx + c$ if $a,b,c$ are negative (as if $x$ was positive, the LHS would be negative). Hence both operations keep everything negative, so we can't hit a positive number. Now we prove everything else works. Suppose something else did not work. WLOG that $v$ is positive. Define "is on the board" to mean there exist operations so that we can write this number on the board. Claim: $-1$ is on the board. Proof: Suppose otherwise. Notice that $u + v$ is on the board and $u, v, u + v$ are pairwise distinct. Consider the quadratic $vx^2 + (u+v) x + u = 0$. This has $-1$ as a root. $\square$ Claim: $1$ and $0$ are on the board Proof: Suppose that $1$ wasn't on the board. Since $v > 1$, we can repeatedly subtract by $-1$ and get that $1$ is on the board, absurd. Hence $1$ must be on the board. Now, $1 + (-1) = 0$ is also on the board. $\square$ Claim: For any integer $c\not \in \{0,1\}$, if $c$ and $c + 1$ are on the board, then $-c$ also is. Proof: Since $1, c, c + 1$ are pairwise distinct and on the board, consider the quadratic \[ 1 \cdot x^2 + (c+1) x + c = 0 \]Clearly $-c$ is a root of this, so $-c$ must be on the board. $\square$ Claim: If some element $r$ is on the board with $|r| > 1$, all integers the same sign of $r$ are on the board. Proof: Suppose otherwise, and let $(-1)^d$ have the same sign of $r$ for some $d\in \{0,1\}$. First we may repeatedly add $r$ by $(-1)^{d+1} $ (since $-1,1$ are on the board) until we get to $2\cdot (-1)^d$. Since we won't hit anything equal to $(-1)^{d+1}$ before this, this implies that $2\cdot (-1)^d$ is on the board. Now for each $k>1$, we can repeatedly add by $(-1)^d$ to get that $k \cdot (-1)^d$ is on the board. Since we know $(-1)^d$ is already on the board, everything of the same sign of $r$ is. $\square$ Since at least one of $u,v$ has absolute value greater than $1$, we see that for some sign, all integers of this sign are on the board. Choose some integer $d$ of this sign with $|d| > 2$. We see that $d\not\in \{0,1\}$, and $d, d+1$ are on the board, so $-d$ must must be on the board. Hence everything of the other sign must also be on the board. Since we had already proven $0$ is on the board, all integers must be on the board.

We are given different integers $u,v$. We consider some cases which arrive naturally while playing with the problem. Case(i):- $u$ and $v$ both are positive, i.e., $u,v>0$ The only first thing we can do is write $u+v$. Then the next obvious step is to make the quadratic $ux^2+(u+v)x+v=0$ Solving it, we get $x=-1$ satisfies the equation and hence we write $-1$ next. Now, we see that adding $-1$ to $u$ yields to us all the positive numbers below $u$ and above $0$, i.e., $u-1, u-2, ...., 0$ but after this we cannot go further since, $0+(-1)=-1$ which cannot be written again as it is already there. Hence, we try to find out $-2$ by making a quadratic. Playing around, we get that $-2$ is the root of the quadratic $ux^2+(3u+v)+2(v+u)=0$. Here, note that any linear combination of $v$ and $u$ can be made and it will be different from the previously present numbers as $u$ and $v$ are positive and distinct. After writing $-2$, we can proceed by adding $-1$ to it and going like this, we can find all the negative numbers. This means that we can write $-(u-1)$ at some point and adding it to $u$, we get that we can write $1$ and hence, by adding $1$ to $u$ and so doing so on, we can also write all the integers greater than $u$. Hence, we get that all pairs $(u,v)$, where $u,v>0$ are are answers. Case(ii):- One of $u,v$ is $0$. This case is important as, if one of $u,v$ is 0, say $u=0$ then we can can only add both the numbers to get $u$ again which we cannot write again and we cannot do anything ahead. Hence, any pair $(k,0)$ or $(0,k)$, where k is an integer is not included in our answer. Case(iii) :- Both $u,v$ are negative, i.e., $u,v<0$. We assume, $u = -a$ and $v = -b$, where $a,b>0$. The only next step is to write $-(a+b)$. Then we make the quadratic $-ax^2-(a+b)x-b=0$ whose root is $x = -1$ which means that we can get all negative numbers less than a(considering $a<b$ => $-a>-b$). Now, we have all the negative numbers written and we want to find whether we can write positive numbers or not. We consider $3$ numbers written $-c, -d, -e$, where $c,d,e>0$. Now, it is evident that by adding $2$ negative numbers we cannot get a positive number and if we make a quadratic $-cx^2-dx-e=0$ which transforms to $cx^2+dx+e=0$, we want a positive root i.e., $x > 0$. But since $c,d,e$ are also positive the quadratic is also always positive and hence it cannot be zero. Hence, we cannot write any positive integers in this case and hence this case is also neglected from our answer. Case(iv):- One of $u, v$ is negative and the other one is positive Here, let $u$ be the positive number and let $v$ be the negative number. We assume $u=a$ and $v=-b$, where $a,b>0$. We follow a similar process as above. If $b > a (a-b<0)$, then, we find the quadratic $ax^2+(a-b)x-b=0$, and find that we can write $-1$. Then, we can write all number less than $a$, i.e., $a-1, a-2, ...., 0$. Then we find the quadratic $ax^2+(3a-b)x+(2a-2b) = 0$, where $-2$ is a root and hence we can write it. And from here we write $-2, -3, ..., -b, -(b+1), ...$. An similar argument follows for where $a>b(a-b>0)$. Lastly in both cases you can write $-(a-1)$ and $1$ and we are done. But a thing to be careful while handling this case is is that the linear combinations can be zero or equal to a previously written number. We need to just check those cases. Those turn out to be when, $b=2a$, $a=2b$ and $a=b$. The first two cases do not affect the process we defined above, i.e., we can reach the value of the coefficients $3a-b$ and $2a-2b$ and the quadratics give the roots $-1$ and $-2$ which are distinct from the number and we can write all integers. But in the third case we get that $(-1,1)$ cannot produce all numbers since adding them gives $0$ adding to which further only leads to $1,-1$ and you cannot reach $3a-b = 2$ - the coefficient of the quadratic and hence only the set ${1,-1,0}$ is produced. Hence, all pairs in this case are solutions except the pair $(1,-1)$. We have exhausted all possible values of the pair $(u,v)$ and summarizing we get the answer to be all $(u,v)$, where $u$ and $v$ are nonzero and atleast one of $u,v$ is positive excluding the pairs $(1,-1)$ and $(-1,1)$.

Solved with SilverBlaze_SY, Jishnu and Oishik. I use $p$ and $q$ instead of $u$ and $v$ respectively. The answer is: $(p,q)$ where $p$, $q$ are both positive. $(p,q)$ where $p\ge 2$ and $q < 0$. $(p,q)$ where $p < 0$ and $q \ge 2$. $(p,q)$ where $p = 1$ and $q < -1$. $(p,q)$ where $p < -1$ and $q = 1$. We first prove that these work. Clearly none of $p$, $q$ can be $ = 0$. For the case $(p,q)$ where $p$, $q$ are both positive, we consider the polynomial $(x+1)(px+q) = px^2 + (p+q)x + q$ which can obviously be obtained by generating $p+q$ from $p$ and $q$. It is easy to note that This gives us $-1$ into our current set of numbers. Thus, now we can go from $p \rightarrow p - 1 \rightarrow \cdots \rightarrow 1$. Now since $p$ and $q$ are distinct, thus one of them must be $> 1$. WLOG be it $q$. Then we can go from $q \rightarrow q + 1 \rightarrow \cdots$ and obtain all the rest of the positive integers. Now we can obtain $-2$ by using $(x + 2)^2 = x^2 + 4x + 4$. Then after that, we can again do $-2 \rightarrow -2 -1 \rightarrow -3 -1\rightarrow \cdots$ and thus we can then obtain all the negative integers too. Now for the case $(p,q)$ where $p\ge 2$ and $q < 0$. We can again use $(x+1)(px+q) = px^2 + (p+q)x + q$ to get $-1$ after which we can obtain the entire set of positive integers from $p$ as above. Now we can do $q \rightarrow q + 1 \rightarrow \cdots \rightarrow -1$. We can also get $-2$ (if not already got, that is $q = -1$) by using $(x + 2)^2 = x^2 + 4x + 4$. After this we can again use $-2 \rightarrow -2 -1 \rightarrow -3 -1\rightarrow \cdots$ to finish. The next $(p,q)$ where $p < 0$ and $q \ge 2$ is analogous to the previous case. For the case $(1,q)$ where $q < -1$. We can first do $q \rightarrow q + 1 \rightarrow \cdots \rightarrow -1$. Then we can just do $-2 \rightarrow -2 - 1 \rightarrow \cdots$ to get the rest of the negative integers. Then we can use $(x-2)(x+1) = x^2 -x -2$ to get $2$ into our current set. Then we can just finish by $2 \rightarrow 3 \rightarrow \cdots$. The other case is also just analogous to this. Now we prove that these cases are the only ones. The other cases are $(1, -1)$, $(-1, 1)$ and $(p,q)$ where $p$, $q$ are both negative. For the first two cases $(1, -1)$ and $(-1,1)$, it is enough to solve it for one of these cases. Note that initially by using only our additional operation, we can get at most to $\left\{-1,0,1\right\}$. After that we do not have any other option other than using the polynomial move. Now we divide the polynomial move into cases, one where $c = 0$ and the other one where $c = \pm 1$ where the polynomial is $ax^2 + bx + c$. Firstly, if $c = 0$, then note that all we can have the roots of the polynomial are $0$ and $\pm 1$. No use. Now for the other case when $c = \pm 1$, by RRT, we can at most produce $\pm 1$ as the integers roots which is of no use either. Thus in this case, we cannot generate all integers. Now for the other case $(p,q)$ where $p$ and $q$ are both negative. Note that by using only the addition operation, we can at most obtain the entire set of negative integers. Now note that if we construct any polynomial with all coefficients that are negative, it is impossible for it to have positive roots. Thus, this must mean we cannot obtain all the integers from this choice of $(p,q)$. Thus we are done.

The pairs $(u,v)=(k,k),(k,0),(0,k)$ can never work because you can't get the third element. Ignore these cases. Replace "written on board" with a set $S$. If $a,b,a+b$ are different integers, then $ax^2+(a+b)x+b=0$ has $x=-1$ as a solution. So, $-1\in S$ is always true if we are ignoring the cases where starting element is $(k,k),(k,0),(0,k)$. Suppose $a>1$. Then, $a-1, a-2, \ldots, 1, 0\in S$. Note that if $1,2\in S$ then $3\in S$, and $4\in S$ and so on. Hence, $\mathbb N\subset S$. Also, $-1\in S$, which means $0\in S$. Hence, equation $0x^2+x+a=0$ for any positive $a\ne0,1$ gives solution $-a$. So, $S=\mathbb Z$. Hence, if $(u,v)$ are integers such that $u\geq1$ and $v\geq1$ but $(u,v)\ne(1,1)$, then we're good. Suppose $-a,-b\in S$ with $a,b>0$ and $a\ne b$. Then, $-1\in S$. So, $-(a+1), -(a+2), \ldots\in S$ which means $\mathbb N_{\leq-a}\subset S$. For any $1\leq c\leq a$, consider equation $(-1)x^2+(-c-a)x+(-ac)=0$. The solutions are $x=-a,-c$. So, $-c\in S$. Hence, $-\mathbb N\in S$. However, at this point the first operation gives ONLY negative numbers while the second operation also gives only negative roots because all three coefficients have same sign! So, we cannot get ANY positive number. If $-a,b\in S$ such that $a,b>0$, then you can just consider $tb-a$ for some large $t$, we eventually get two positive numbers. This however doesn't work if $b\mid a$. In that case, note that $-1\in S$ holds anyways so $b-1,\ldots,1\in S$ and we're done (unless $(u,v)=(1,-1)$, in which case you can't proceed.) So, the pairs that DON'T work are: $(k,k),(k,0),(0,k),(-s,-t)$ where $k\in\mathbb Z$ and $s,t\in\mathbb N$. Rest every pair works.

We start by removing trivial edge cases. It is not possible to generate $\mathbb{Z}$ if: (1) $uv = 0$ (2) $u$ and $v$ are both negative Assume $uv \neq 0$ and that, there is at least one positive number WLOG, $u > v$ (this implies that $u$ is positive) If $u>v>0$, we can generate all sufficiently large multiples of $gcd(u,v)$, and thus we can make all negative integers with operation (ii). Thus, it works. If $v<0$, Note that we can make $-1$ by plugging in $(a,b,c) = (u,u+v,v)$. Thus, we can generate all numbers between $u$ and $v$. If $v < -1$ we can generate all negative numbers $\Rightarrow$ works If $v = -1$ and $u > 1$, we can generate all positive integers $\Rightarrow$ works If $(u,v) = (1,-1)$, we can see it doesn't work Thus, conditions for not working are: (1) $uv = 0$ (2) $u$ and $v$ are both negative (3) $uv = -1$

Hopefully, I got it.

WLOG let $u\geq v$. With the first operation, we get $u+v$. By using the second operation as $a=u, b=u+v, c=v$, we get $-1$. So, it means that if $u>1$, we can get any integer at most $u$. Then we have $1$, which means we can get any integer at least $u$. If $u=1$, and $v\leq-2$, we can get $-1$, and using the the second operation as $a=1, b=-1, c=-2$, we get $2$. The rest is obvious. If $u=1$, and $v=-1$, we get $0$. It's pretty easy to see that no matter how we use the second operation, we cannot get a new number. If $u,v<0$, let's say we can get a positive number $x$ after some finite number of operations. We can only get the that positive number using the second operation because adding two negative number gives you a negative number. By the quadratic formula, $\hspace{6cm}$$ \begin{array}{*{20}c} {x = \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}}} >0 \\ \end{array}$ $\sqrt {b^2 - 4ac}$ and $-b$ are positive,$-b\geq\sqrt {b^2 - 4ac}$ and $-2a$ is negative. So it means, $x<0$, contradiction.

Solution from Twitch Solves ISL: Note that the following cases are impossible: If either $u = 0$ or $v = 0$, operation (i) cannot produce numbers, and (ii) cannot be used at all. Hence the task is obviously impossible. If $\{u,v\} = \{-1,1\}$, operation (i) produces $0$, and no more numbers can be added with operation (i) or (ii). So this case cannot be done. If $u,v < 0$, then we claim all numbers written are negative. Indeed, (i) respects this; and also, if $a,b,c < 0$ then no $x > 0$ can obey $ax^2+bx+c=0$. So the task is impossible here too. We claim the task is possible in all other cases. Claim: We can write $-1$. Proof. We can write $u+v$, and the three numbers $u$, $v$, $u+v$ are different. The quadratic $ux^2 + (u+v)x + v$ has root $-1$. $\blacksquare$ Claim: We can write a positive number $m \ge 2$ on the board. Proof. We assumed that at least one of $u$ and $v$; say $u = \max(u,v) > 0$. If $u \ge 2$ we're done. Suppose $u = 1$; then $v$ is some negative number with $v \le -2$. We can write $(-1) + 1 = 0$. Then $0x^2 + x + v = 0$ has root $-v$, which is the desired number. $\blacksquare$ Let $m \ge 2$ be from the claim; then we can write $m-1 \ge 1$, and hence we can write all of $m-1$, $2m-1$, $3m-1$, \dots. Thus (because we have $-1$) we can then write any nonnegative integer by adding $-1$ several times. (However, we cannot write $-2$ in this way, because we cannot add $-1$ to itself.) But now for $n > 0$, $-n$ is a root of $0x^2 + x + n$, so we can write the negative integers too.

Claim - All integers can be written on the board iff - 1. $u \neq 0, v \neq 0$ 2. At least one of $u, v$ is positive. 3. $u$ and $v$ are not $-1$ and $1$. Proof - (If part) WLOG assume $u>v$. Case 1 ($u \neq -v$) - Assume we have integers $u$ and $v$ satisfying the given conditions. Then, we can obtain any integer of the form $k_1u+k_2v$ such that $k_1,k_2 \in \mathbb N$, by simply adding $u$ and $v$ the required number of times (from Operation (i)). This allows us to obtain $u+v$, $2u+2v$ and $3u+3v$. Using Operation (ii), we can write the roots of $(u+v)x^2 + (3u+3v)x + 2u+2v = 0$, which are $-1$ and $-2$. We know that one of $(u, v)$ is positive, which means that $u$ is positive. If $u>1$, then using $-1$ and $u$, we can get all integers from $0$ to $u$ (which includes $1$), and using $1$ and $u$, we can get all integers $>u$. Using $-2$ and $-1$ gives us all negative integers. If $u=1$, then $v<-1$ (since $v \neq 0, -1$ and $v<u$). Using $-2$ and $-1$, we get all negative integers, and using $-1$ and $1$, we get 0. Afterwards, using Operation (ii) on $1$, $0$ and $-k^2$ gives us one root as $k$ ($\forall k \in \mathbb N$). Since we have all negative integers, it is possible to obtain all naturals. Thus, all integers can be obtained. Case 2 ($u = -v$) - We have $(u, -u)$ such that $u > 1$. Adding them gives $0$. Using Operation (ii) on $u$, $0$ and $ -u$ gives the equation $ux^2 - u = 0$, which has roots $\pm 1$. Using $u$ and $-1$ gives all integers from $0$ to $u$, and using $u$ and $1$ gives all integers $>u$. Similarly, using $ -u$ and $1$ gives all integers from $0$ to $ -u$, and using $ -u$ and $-1$ gives all integers $<-u$. Thus, all integers can be obtained. (Only if part) If we have the initial pair $(u,0)$, then adding gives us $u$, which is already written on the board. Operation (ii) cannot even be used since there are only $2$ integers on the board. So, $(u,0)$ is not possible. If both $u$ and $v$ are negative, then using Operation (i) always gives a negative integer. Also, if all coefficients in a quadratic equation are negative, it gives $-b/a = $sum of roots is negative and $c/a = $product of roots is positive, implying that both roots are negative. This means that even Operation (ii) will only give negative numbers. So we will never be able to write a positive integer on the board. If $(u,v)$ is $(1,-1)$, adding gives us $0$. Adding any 2 of $(1,0,-1)$ will only result in $(1,0,-1)$. Using Operation (ii), we can form equations $x^2 - x = 0$, $x^2 - 1 = 0$, $x - 1 = 0$, all of which have roots only in $(1, 0, -1)$. So we can never write all integers. Hence, the claim is true.

We claim that only pairs $(u,v)$ doesn't work iff $(u,v)=(k,-k)$ for some integer $k$ and $u,v$ are both negatives. $(u,v)=(k,-k)$ is obvious, let us prove for $u$ and $v$ are negatives. Suppose $a$ and $b$ are both negatives.So $a+b$ is also negative. Now suppose $a,b,c$ are $3$ negatives. There exist some $x$ such $ax^2+bx+c=0$ Suppose $x$ is positive. Now $ax^2$ is negative, $bx$ is negative as $x$ is positive and $c$ is also negative. So sum of 3 negatives can't be 0 which is a contradiction. Now we will prove other pairs $(u,v)$ work. \(Lemma:\) $(1,k)$ works where $k$ is a positive integer. \(Proof:\) We can keep adding $1$ and get all integers $>k$. Now For any $(u,v)$, $ux^2+(u+v)x+v=0$ has solution $x=-1$. We can get all integers less than $k$ Now for any $(u,v)$, if one of them is greater than $1$. We can get $1$ and reach all integers. $\square$

Answer is $(u,v)$ where at least one is positive and the other is nonzero, and $(u,v)\neq(-1,1)$ and $(1,-1)$

The answer is all $(u, v)$ such that at least one is positive, none are zero, and they are not a permutation of $(1, -1)$. It is easy to see why these don't work. Namely, both negatives cannot reach positives, zero only makes two numbers, permutations of $(1, -1)$ gets stuck after making a zero. As a first observation, we can use $u, v, u+v$ to form $-1$ ALWAYS. Now we tackle the case where one is negative and one is positive (in fact this is the only case we need to consider because we can keep subtracting $-1$). Let $u \geq 1$ and $v \leq -1$, where $|u| \geq |v|$. Note that if this last condition is not satisfied, we can keep adding $u$ to $v$ until it does. Then, keep subtracting $1$ from $u$ until we reach $-1$. Within this process we formed $1$. Next we claim that we can form a number greater than or equal to two. This is not immediate only if $u=1$. Then $v \leq -2$, and we can form $-v$ by taking $a=0, b=1, c=v$ which finishes. Now, the problem is trivial because everything above $u$ we can find by adding $1$ and everything below $v$ we can subtract $-1$. We can get the pesky numbers after $1$ using our number greater than or equal to two now we are done.