We will prove the answer are all pairs such that $(m,n)=1$ If $d=gcd(m,n)>1$ then the set of all multiples of $d$ satisfy the question too, hence this pair isn't one we want. If $d=1$ then by $x=y=m$ we can get all multiples of $m^2$ and similar for $n^2.$ Because $gcd(m^2,n^2)=1$ there exist $a,b$ such that $am^2-bn^2=1$ We can form $1$ with $x=am^2, y=bn^2$ and $k=-2$ . After that is is trivial the only admissable set containing $1$ is the one of all integers.

SCP wrote: We will prove the answer are all pairs such that $(m,n)=1$ If $d=gcd(m,n)>1$ then the set of all multiples of $d$ satisfy the question too, hence this pair isn't one we want. If $d=1$ then by $x=y=m$ we can get all multiples of $m^2$ and similar for $n^2.$ Because $gcd(m^2,n^2)=1$ there exist $a,b$ such that $am^2-bn^2=1$ We can form $1$ with $x=am^2, y=bn^2$ and $k=-2$ . After that is is trivial the only admissable set containing $1$ is the one of all integers. your solution is same with mine^^ I think it's a good, simple question.

This is ridiculous: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=151&t=507200&p=2849284&hilit=X%5E2+kxy+y%5E2#p2849284

It is very easy to conjecture $(m,n)=1$, because if $(m,n) \textgreater 1$, it obviously doesn't work ($A = (m,n) \times \mathbb{Z}$), and it is very difficult to come up with a counterexample if $(m,n)=1$. And whenever coprime integers and $\mathbb{Z}$ are involved, one can smell Bezout. However, the condition is a quadratic one, not a linear one, so the trick is to surpass this obstacle, by introducing some expressions of degree $2$ that belong to $A$. Now, assume $(m,n)=1$ and we have $m,n \in A$ an admissible set. We will try to prove $A = \mathbb{Z}$. Take any $z \in A$. Clearly with $x=y=z$, $kz^2 \in A \forall k$. So if $z=1 \in A$, we're done. So we'll try to find $x,y \in A$ such that $x^2+kxy+y^2=1$. If $k=-2$, we can easily factor this and it will suffice to find $x, y \in A$ such that $x-y=1$. So if we find two consecutive integers, both in $A$, we're done. But remember that $am^2 \in A$ and $bm^2 \in A$ for all $a, b$ integers. By Bezout, we can set $a$, $b$ such that $am^2-bn^2=1$, and we have found two consecutive integers, both in $A$. So we're done.

We will prove that for all integers such that $(m,n)=1$.It is obvious that if $(m,n)=d>1$ then every element of the set $A$ is divisible with $d$ and the set $A$ is ${...-3d,-2d,-d,0,d,2d,3d....}$ satisfayes the conditions.Now,if $(m,n)=1$,then plugg $x=y=m$ and $x=y=n$ that every integer of the form $k*m^2$ and $k*n^2$ is in the set,so we can pick $a$ and $b$ such that $a*m^2-b*n^2=1$(by chineese remainder theorem),so plug $x=a*m^2$ and $y=b*n^2$ and $k=-2$we obtain $1$,so now it is trivial that we have all integers.

Answer: If $(m,n)=1$, the admissible set containing both $m$ and $n$ is the set of all integers. Let's think $(m,n)=g>1$. Then every element of $A$ is divisible by $g$. Because $g$ is underneath every integer there. So, the set $\{...,-3g,-2g,-g,0,g,2g,3g,....\}$ will work. Now we'll prove that if $(m,n)=1$ the admissible set containing both $m$ and $n$ is the set of all integers. Lemma 1: If we have $j \in A$, then every multiple of $j^2$ will be in $A$. Proof: By taking $x=y=m$ we will have $m^2+km^2+m^2=2m^2+km^2=m^2(k+2)$. Then by putting $k= \{-2,-1,0,1,2\}$ we will have every multiple of $j$. Now the main part. We've $(m,n)=1$. So, $(m^2,n^2)=1$. By bezout's identity there exists $a,b$ such that $am^2+bn^2=1$. Now by applying lemma 1 if we've $m,n$ in $A$ then we will also have $am^2$ and $bn^2$ in $A$. So by taking $x=am^2, y=bn^2, k=2$ we have $$(am^2)^2+2(am^2)(bn^2)+(bn^2)^2= (am^2+bn^2)^2=1$$So, we can always form $1$ by taking these values from a set whare $(m,n)=1$. Then by taking $x=y=1$ and taking $k= \{...,-2,-1,0,1,2,...\}$ we will have every integer in $A$. $\mathbb Q. \exists .\mathbb D.$

We claim the $(x, y)$ which satisfy the problem condition are precisely those for which $\gcd(x, y) = 1$. We first show that these pairs work. Lemma: The elements $x, y, x^2 + y^2, x^2 + xy + y^2 \in A$ are pairwise relatively prime. Proof: By assumption $\gcd(x, y) = 1$. Also, $\gcd(x^2 + y^2, x) = \gcd(y^2, x) \leq \gcd(y, x) = 1$, implying that $\gcd(x^2 + y^2, x) = 1$. Similarly, $\gcd(x^2 + y^2, y) = 1$. Additionally, $\gcd(x^2 + xy + y^2, x^2 + y^2) = \gcd(xy, x^2 + y^2) \leq \gcd(x, x^2 + y^2) \cdot \gcd(y, x^2 + y^2) = 1$, so $\gcd(x^2 + xy + y^2, x^2 + y^2) = 1$. Finally, $\gcd(x^2 + xy + y^2, x) = \gcd(y^2, x) \leq \gcd(y, x) = 1$, so $\gcd(x^2 + xy + y^2, x) = 1$. Similarly, $\gcd(x^2 + xy + y^2, y) = 1$, proving the lemma. $\blacksquare$ Now, we have that all $a \equiv x^2 + y^2 \pmod{xy}$ are members of $A$, and additionally that all $a \equiv (x^2 + xy + y^2)^2 + (x^2 + y^2)^2 \pmod{(x^2 + xy + y^2)(x^2 + y^2)}$ are members of $A$. Since $\gcd(xy, (x^2 + xy + y^2)(x^2 + y^2)) = 1$, by CRT there exists $N$ such that \begin{align*} N &\equiv x^2 + y^2 \pmod{xy} \\ N + 1 &\equiv (x^2 + xy + y^2)^2 + (x^2 + y^2)^2 \pmod{(x^2 + xy + y^2)(x^2 + y^2)}. \end{align*}For such an $N$, both $N$ and $N + 1$ are members of $A$. Now, $N^2 + (N + 1)^2 - 2N(N + 1) = 1 \in A$. To conclude, $1^2 + 1^2 + 1 \cdot 1 \cdot k = k + 2 \in A$ for all $k \in \mathbb{Z}$, implying that $A$ contains all integers. Thus, the pairs $(x, y)$ for which $\gcd(x, y) = 1$ satisfy the problem condition. Finally, if $\gcd(x, y) = d > 1$, the set $A$ of all multiples of $d$ is an admissible set which contains $x, y$, but $A \neq \mathbb{Z}$. Thus, the only pairs $(x, y)$ which work are those described initially, so we are done. $\Box$

The answer is all integers such that they are relatively prime. First we set $k=2$ to get that $(x+y)^2 \in \mathcal{A}$, then we set $k=-2$ to get that $(x-y)^2 \in \mathcal{A}$, also we have that when $k=0$ that $x^2+y^2 \in \mathcal{A}$. Setting $x=y$ we get that $0 \in \mathcal{A}$, this implies that $x^2 \in \mathcal{A}$, this means that if $x^2 \in \mathcal{A}$, then we have that $x \in \mathcal{A}$. This means that $x+y \in \mathcal{A}$ and $x-y \in \mathcal{A}$, this implies that for every integer $k$ we have that $kx \in \mathcal{A}$, thus we need to get that $x=1$. We use the following ultra well known lemma: Lemma: Let $d=(x,y)$, then there exist some integer numbers $k$ and $q$ such that $kx+qy=d$. By this lemma we easily get that $d=(x,y) \in \mathcal{A}$ Now if $d=1$ we win and $\mathcal{A}=\mathbb{Z}$ If $d \neq 1$ , then we have that we only generate pairs of numbers such that they are divisible by $d$, thus $\mathcal{A} \neq \mathbb{Z}$

Solved with nukelauncher. Clearly, $\gcd(m,n)=1$ since $\gcd(m,n)$ divides every element of $A$. We claim all such pairs work. Let $P(x,y,k)$ denote the assertion $x^2+kxy+y^2\in A$. By plugging in $(x,y)=(m,m)$, we get that $am^2 \in A$ for any $a$, and similarly $bn^2 \in A$ for any $b$. Since $\gcd(m,n)=1$ implies $\gcd(m^2,n^2)=1$, we can find $a,b\in\mathbb Z$ such that $am^2-bn^2=1$. Plugging in $(x,y,k)=(am^2,bn^2,-2)$ gives \[ 1=(am^2-bn^2)^2 \in A. \]Finally, $(x,y,k)=(1,1,k)$ gives $2+k \in A$ for any $k$, i.e. $A=\mathbb{Z}$.

The same idea was used in USAMO 2004/2

I claim the answer is that this is true for all pairs such that $\gcd(m,n)=1$. Clearly if $\gcd(m,n)=g>1$, then any new element satisfies \[g^2 \mid x^2+kxy+y^2\]Thus, we cannot generate the necessary infinitely many elements not divisible by $g^2$. We now construct a solution for all $\gcd(m,n)=1$. Note that by plugging in $(m,m)$, we get all multiples of $m^2$, similarly we can get all multiples of $n^2$. Since $\gcd(m,n)=1\Longrightarrow \gcd(m^2,n^2)=1$, we have that by Bezout's we can find $x,y$ such that $xm^2-yn^2=1$. Thus, we can plug in $(xm^2,yn^2)$ with $k=2$ which yields \[(xm^2-yn^2)^2 = 1\] Thus 1 is in the set, and by plugging in $x=y=1$ we get all integers.

Answer. $\gcd (m,n)=1$. Proof. Notice that all non-empty admissible sets contain $0$ if we take $x=y,k=2$. If $\gcd(m,n)=d>1$, then the set $d\mathbb{Z}$ indeed satisfies the condition. If $\gcd(m,n)=1$, then take $x=y=m$ and arbitrary $k$ $\implies m^2\mathbb{Z} \subseteq A$, similarly $n^2\mathbb{Z} \subseteq A$. Since $\gcd(m^2,n^2)=1$, by Bezout's identity there exist $a,b \in \mathbb{Z}$ such that $am^2+bn^2=1$. Then take $x=am^2,y=bn^2,k=2 \implies 1 \in A$. Now take $x=y=1 \implies A \equiv \mathbb{Z}$. $\blacksquare$

These are all pairs $(m,n)$ s. t. $\gcd(m,n)=1$. Others won't work because if the gcd is $d>1$, then the set of integers divisible by $d$ satisfies the condition as well. To prove that they work, notice that we can get $lm^2\in A$ for all integers $l$ by choosing $x=y=m$ and $k=l-2$. By choosing $x=m$, $y=n$ and $k=-2$, we can get $(m-n)^2\in A$ as well. As $m$ and $n$ are coprime, by Bézout, there exist integers $k,l$ s. t. $$(m-n)^4+kn(m-n)^2+n^2=lm^2+1$$But notice that $A$ contains the LHS because $(m-n)^2\in A$ and $n\in A$. Also, $lm^2\in A$. So we have to integers $x,y\in A$ that differ by 1. Taking $k=-2$, we get that $1=(x-y)^2\in A$, but then taking $x=y=1$ gives $k+2\in A$ for all integers $k$, as desired.

Solution from Twitch Solves ISL: The answer is $\gcd(m,n) = 1$. If $\gcd(m,n) > 1$, one can just let $A$ be multiples of $\gcd(m,n)$. On the other hand, suppose $\gcd(m,n) = 1$. Let $P(x,y,k)$ be the statement. Then: $P(m,m,k)$ and $P(n,n,k)$ show all multiples of $m^2$ and $n^2$ are in $A$. $P(am^2,bn^2,2)$ gives \[ a^2 \cdot m^4 + 2ab \cdot m^2n^2 + b^2 n^4 = (am^2+bn^2)^2 \in A \]which shows every perfect square is in $A$. In particular $1 \in A$. Now $P(1,1,k)$ implies $A = {\mathbb Z}$.

Basically same sol. We claim the answer is iff $\gcd(m,n)=1$. Clearly, if $d = \gcd(m,n) \not = 1$ then all elements in $A$ are divisible by $d$ which cannot make it the set of integers. Now assume $\gcd(m,n) = 1$. We have $m^2+kmm+m^2 = (k+2)m^2 \in A$ for all $k$ which yields all multiples of $m^2$. Similarly, all multiples of $n^2$ is in $A$. Recall $\gcd(m^2,n^2)=1$ so, by Bezout's there exists $a,b \in \mathbb{Z}$ where $am^2+bn^2 = 1$. We have $am^2, bn^2 \in A$ so $(am^2)^2+2(am^2)(bn^2)+(bn^2)^2 = (am^2+bn^2)^2 =1 \in A$ Applying the operation to $1$ and $1$ yields $1^2+k(1)(1)+1^2 = k+2 \in A$ for all $k$. Which takes on all integer value. $\blacksquare$

We claim that the answer is all $(m,n)$ such that $m$ and $n$ are relatively prime. Notice that if $\text{gcd} (m,n)=d>1$, then the set of all multiples of $d$ is an admissible set, not the set of integers, that contains $m$ and $n$, so we focus on proving that if $m$ and $n$ are relatively prime, the only admissible set containing $m$ and $n$ is the set of integers. Suppose we have an admissible set containing relatively prime $m$ and $n$. Plugging in $x=y=m$, we get that any multiple of $m^2$ should be in the set. Similarly, any multiple of $n^2$ should be in the set. Consider the following well-known lemma: Lemma: If $\text{gcd} (x,y)=1$, there exists a multiple of $x$ and a multiple of $y$ that are one apart. Applying this lemma to the problem, because $\text{gcd} (m^2,n^2)=1$, there exist a multiple of $m^2$ and a multiple of $n^2$ that are one apart, and therefore two integers in this set, $r$ and $s$, are one apart. Simply taking $r^2-2rs+s^2=(r-s)^2$ tells us that $1$ must be in the admissible set, and then plugging $x=y=1$ finishes. We are done.

We claim the answer is all integers $m,n$ such that $(m,n)=1$. Let $f(a,b,k)=a^2+b^2+kab$. First note that if $(m,n)=d> 1$, all elements generated from $m,n$ are multiples of $d$, so the set generated by $m,n$ can't be all integers. Now we prove that those pairs work. Note that $f(a,a,\ell - 2)$ implies that, for every $a\in A$, every multiple of $a^2$ is also in $A$. Then, since $1=(m,n)=(m^2,n^2)$, by Bézout, there are integers $x,y$ such that $xm^2+yn^2=1$. Hence, $f(xm^2,yn^2,2)$ proves that $1\in A$. Therefore, $f(1,1,\ell-2)$ implies that $A=\mathbb Z$. $\blacksquare$

We claim the answer is $\boxed{\gcd(m,n) = 1}$. If $\gcd(m,n) \ne 1$, then let $p$ be a prime dividing $m,n$. The set of all integer multiples of $p$ is admissible, a contradiction. Therefore, $\gcd(m,n) = 1$. Now we prove that for all coprime $m,n$, the only admissible set containing $m,n$ is the set of all integers. Setting $x = y$ gives that if $x\in A$, then any integer multiple of $x^2$ is also in $A$. Thus, all multiples of $m^2$ and all multiples of $n^2$ are in $A$. By Bezout, there exists an $a$ where $a, a + 1$ are in $A$. Setting $x = a, y = a + 1$, and $k = -2$ gives $a^2 + (a+1)^2 - 2 a(a+1) = (a - (a+1))^2 = 1 \in A$, so all multiples of $1^2 = 1$ are in $A$, implying all integers are in $A$.

We claim the answer is all $(m,n)$ such that $\gcd(m,n)=1.$ If $d=\gcd(m,n)\neq1,$ we know $d\mid x^2+kxy+y^2$ so all integers $\ell$ such that $d\nmid\ell$ cannot be in $A.$ Suppose $d=1.$ Letting $x=y=m$ yields $(k+2)m^2\in A$ and similarly $(k+2)n^2\in A.$ Hence, all multiples of $m^2$ and $n^2$ are in $A.$ By Bezout's lemma, $am^2-bn^2=1$ for some $(a,b).$ But $$(am^2)^2+(bn^2)^2-2(am^2\cdot bn^2)=(am^2-bn^2)^2=1\in A.$$Therefore, $1^2+k+1^2=k+2\in A$ and all integers are in $A.$ $\square$

We claim the answer is all $(m,n)$ such that $\gcd(m,n)=1$. If $\gcd(m,n)=d>1$, then the set of all multiples of $d$ works. If $\gcd(m,n)=1$, then note that $(k+2)m^2 \in A$ and $(k+2)n^2 \in A$. Note that $\gcd(m^2,n^2)=1$, so by Bezout's, there exists integers $a$ and $b$ such that $am^2-bn^2=1$. Then taking $k=2$, $x=am^2$, and $y=bn^2$, which are in the set by the above, we get that $$(am^2-bn^2)^2 \in A \implies 1 \in A.$$Thus $k+2 \in A$ for all integers $k$, and we are done.

Note that $A=\{x\gcd(m,n)\mid x\in\mathbb N\}$ is admissible and contains $m$ and $n$. If $\gcd(m,n)>1$ then this is not the set of all integers. So we only consider now when $\gcd(m,n)=1$. Since $m\in A$ then $km^2\in A$ for integers $k$ (by setting $x=y=m$), similarly $kn^2\in A$. Since $\gcd(m^2,n^2)=\gcd(m,n)^2=1$, there are integers $a,b$ with $am^2-bn^2=1$ (by Bezout's Lemma). Then: $$a^2m^4-2am^2bn^2+b^2n^4=(am^2-bn^2)^2=1\in A.$$Finally, setting $x=y=1$ we have $k+2\in A$ for integers $k$, so $A=\mathbb Z$.

The answer is all $m,n$ which satisfy $\gcd{(m,n)}=1$. Clearly when $\gcd{(m,n)}>1$ we can take the set which consists of all multiples of $\gcd{(m,n)}$. Otherwise, note that all multiples of $m^2$ and $n^2$ are contained in the set. We also have $\gcd{(m^2,n^2)}=1$ so there exist $a,b$ such that $am^2-bn^2=1$ and since $am^2$ and $bn^2$ are in the set we have that \[(am^2)^2-2(am^2)(bn^2)+(bn^2)^2=1\]is in the set. But now all multiples of $1^2=1$ are in the set. We are done. $\blacksquare$

Call an unordered pair $m,n$ good if the only admissible set with $m$ and $n$ is $\mathbb{Z}$ and call an integer $a$ good if the only admissible set with $a$ is $\mathbb{Z}$ Note that if $a \in A$ where $|a|=1$ then $a^2+a^2+ka^2=2+k\in A$ for all integers $k.$ Thus, $1$ and $-1$ are good. More generally, if $a\in A$ then $a^2(2+k)\in A.$ Thus, all multiplies of $a^2$ can be made. If $(a,b)=1$ then $(a^2,b^2)=1$ so by bezout's identity we can add $x,y$ two multiples of $a^2,b^2$ respectively and obtain $1$. Letting $k=2$ we see that $(x+y)^2=1\in A.$ If $(m,n)\neq 1$ simply take all integers divisible by $(m,n).$ Thus, the only possible $m,n$ are the coprime ones.

Clearly, if $m$ and $n$ are not relatively prime, the set of all multiples of $\gcd(m,n)$ would work, so $m$ and $n$ have to be relatively prime. We can generate all multiples of $n^2$ and $m^2$ by using $x=y=n$ and $x=y=m$. Note that if $m$ and $n$ are relatively prime, by the Chinese Remainder theorem, there exists $s$ such that $s\equiv 0 \pmod{n^2}$ and $s\equiv -1 \pmod{m^2},$ so $s$ and $s+1$ are both in the set $A$. Now, by using $x=s$ and $y=s+1$ and choosing $k=-2$, 1 is in set $A$. Using $x=y=1$, we can then generate all integers. Therefore, the answer is all $(m,n)$ such that $m$ and $n$ are relatively prime.

what is the motivation for thinking bezout theorem

The answers are $m,n$ which are relatively prime. First, let's consider when $\gcd(m,n)=d>1$. Note that the set that contains all (and only) multiples of $d$ is admissible. This is because for any $x,y$ in the set, the value of $x^2+kxy+y^2$ is divisible by $d$ since both $x,y$ are divisible by $d$. But this set doesn't cover all integers, e.g. $1$. Hence, we get a contradiction. Now, suppose that $m,n$ are relatively prime. We make the following observation. If $x,y\in A$ then $(x-y)^2=x^2-2xy+y^2\in A$. If $a(,a)\in A$ then $a^2b=a^2+(b-2)(a\cdot a)+a^2\in A$ where $b$ is any integers. In particular, if $1\in A$ then the conclusion follows. Therefore, we want to prove that $1\in A$. Using the second observation, we have that $am^2, bn^2\in A$ for any integers $a,b$. Note that by Bezout's Lemma, there exist $a,b$ such that $$am^2-bn^2=1.$$And so using the first observation, we get that $(am^2-bn^2)^2=1\in A$, and we're done.

The answer is all pairs $(m,n)$ of coprime numbers. It's obvious that set $A^*=\gcd (m,n)\times \mathbb{Z}$ is the admissible set containing $m,n,$ so all suitable pairs are coprime. On the other hand, we may put $x=y=m$ and $x=y=n$ getting that any admissible set $A\ni m,n$ contains all multiples of $m^2,n^2.$ In particular, $\exists \alpha m^2,\beta n^2\in A:\alpha m^2-\beta n^2=1.$ Substituting $\alpha m^2=x,\beta n^2=y,k=-2$ we get $1\in A.$ Finally, substituting $x=y=1,k=z-2$ we obtain that $z\in A$ for any $z\in \mathbb Z,$ so the conclusion follows.

The answer is when $\boxed{\gcd(m,n)=1}$. Obviously this is necessary as otherwise everything in $A$ has to be a multiple of $\gcd(m,n)$. Now we will prove that this is sufficient. By Bezout's, there exists integers $a$ and $b$ such that \[ax^2-by^2=1.\]Now let $k_1=(2y^2-2x^2+1)a$ and $k_2=(2y^2-2x^2+1)b$. Then \[m=x^2+k_1x^2+x^2=((2y^2-2x^2+1)a+2)x^2\in A\]and \[n=y^2+k_2y^2+y^2=((2y^2-2x^2+1)b+2)y^2\in A.\]But $m=n+1$, so \[m^2+(-2)(m)(n)+n^2=(m-n)^2=1\in A.\]Now for any positive integer $z$ \[(1)^2+(z-2)(1)(1)+(1)^2=z\in A\,\blacksquare\]

All coprime pairs. If not, just take a prime ideal containing both. If $m, n$ are coprime, then all multiples of $m^2, n^2$ lie in the set from $(m, m)$ and $(n, n)$. Thus by Bezout's theorem, there exists $s, t \in A$ such that $s - t = 1$. However by $k = -2$ we must have $1 = (s-t)^2 \in A$, then from $(1, 1)$ we have all integers must lie in $A$.

All integers $(m, n)$ such that $\gcd(m,n) = 1$ work. Clearly $m,n$ not satisfying this condition do not work; then all numbers in $A$ must be a multiple of $\gcd(m,n)$, contradiction. Otherwise, $(x, y) = (m,m)$ and $(x,y) = (n,n)$ show that all multiples of $m^2, n^2$ are in $A$; by Bezout's, there exists integers $a,b$ so that \[am^2 - bn^2 = 1;\]taking $(x,y) = (am^2, bn^2)$ and $k = -2$ shows that $1 \in A$; now it is trivial to check that $(1, 1)$ gives that all integers are in $A$.

Nice problem: First note that if $(m,n)=(1,1)$ or $(m,n)=(-1,-1)$ results in $A=\mathbb{Z}$, as $1^2+k\cdot1\cdot1+1^2=2+k$ covers all the integers. Now, let $\gcd(m,n)=s>1$ so that $m=xs$ and $s=ys$. Then each element of $A$ is divisble by $s^2$, so in particular, $1\notin A$, and thus $A\neq\mathbb{Z}$. Claim: If $m,n\in A$ and $\gcd(m,n)=1$, then $A=\mathbb{Z}$. Proof: By taking $m=m$, $m^2+km^2+m^2=(k+2)m^2$ all multiples of $m^2$ are in $A$. Similarly, all multiples of $n^2$ are in $A$. By Bezout's theorem, there exist integers $x,y$ such that $xm^2-yn^2=1$. So there are two elements in $A$ which are consecutive. Let them be $r$ and $r+1$. Then, \[r^2-2r(r+1)+(r+1)^2=r^2-2r^2-2r+r^2+2r+1=1\in A\]and thus, by our first observation, $A=\mathbb{Z}$, as desired.

$\color{magenta}\boxed{\textbf{SOLUTION N1}}$ If $\gcd(m,n) > 1$, Then $A =$ the multiples of $\gcd(m,n)$ works So, Suppose $\gcd(m,n) = 1$. Let $P(x,y,k)$ be the statement. Then, $\bullet P(m,m,k)$ and $P(n,n,k)$ $\implies$ all multiples of $m^2$ and $n^2$ are in $A$. $\bullet P(am^2,bn^2,2)$ gives $$a^2 \cdot m^4 + 2ab \cdot m^2n^2 + b^2 n^4 = (am^2+bn^2)^2 \in A$$Hence, $1 \in A$. $\bullet P(1,1,k) \implies A = {\mathbb Z} \blacksquare$.

We claim the answer is for all pairs such that $gcd(m,n)=1$ Assume that $gcd(m.n) > 1$, and assume that $gcd(m,n)=d$ then we can see that the set $A$ will be equal to the multiples of $d$, namely $$A= \{...-3d,-2d,-d,0,d,2d,3d,....\}$$ Now we show that $gcd(m.n)=1 $ works. Claim 1: If $j$ is in $A$ then all multiples of $j^2$ are as well in $A$: Plugging in $(m,m)$ in the equation \[x^2+kxy+y^2 (1) \]We will have all multiples of $m^2$, similarly for $n^2$. Claim 2: If $gcd(m,n)=1 $ then $1 \in A$: We know that $gcd(m^2,n^2)=1$. Here we apply Bezout lemma, so there exists $a,b$ such that $am^2+bn^2=1$, so plugging $x=am^2$,$y=bn^2$ and $k=-2$ we have : $$(am^2)^2+2(am^2)(bn^2)+(bn^2)^2= (am^2+bn^2)^2=1$$Meaning that $1 \in A $ as desired. Now by Claim 1 we have that all multiples of $1$ are in $A$, so $A$ is the set of all integers.

Clearly if $m$ and $n$ are not relatively prime then all elements in $A$ will be divisible by $\gcd(m, n)$ which is undesirable(since we won't be able to get numbers that are not divisible by $\gcd(m, n)$. So then $\gcd(m, n) = 1$. We can plug in $(m, m)$ and $(n, n)$ and vary $k$ to get multiples of $m^2$ and $n^2$ respectively. Since Bezout's lemma states there exists $a$ and $b$ so that $ax + by = \gcd(x, y)$ we can plug in $(am^2, bn^2)$ with $k = 2$ to get $(am^2 + bn^2) \in A$ with $a$ and $b$ chosen so that $am^2 + bn^2 = 1$. Now that $1 \in A$ we can plug in $(1, 1)$ and vary $k$ to get all integers, done.

We claim $\boxed{\gcd(m,n)=1}$ is necessary and sufficent. It is necessary since the admissible set generated by $m$ and $n$ is a subset of $(m,n)$ (the ideal). Suppose $\gcd(m,n)=1$. Let $A$ be the admissible set generated by $m$ and $n$. Then $xm^2,yn^2\in A$ for all integers $x$ and $y$. By Bezout, there exists integers $x$ and $y$ such that $xm^2+yn^2=1$. Then $(xm^2)^2+2(xm^2)(yn^2)+(yn^2)^2=1$ so $1\in A$. The conclusion follows. $\square$

I claim the answer is only the pairs $m,n$ with greatest common divisor $1$. To prove nothing else works, note that the set of all numbers divisible by the greatest common divisor of $m,n$ is a working solution. To prove all such pairs work, consider all numbers of the form $2m^2 + km^2, 2n^2 + kn^2$ are in the set. We can then find two of these numbers with difference $1$ by Bezout's theorem, then set $k = - 2$ and we get that $1$ is part of the set. We are clearly done from there, as all numbers of the form $1^2 + k \cdot 1 \cdot 1 + 1^2 = 2 + k$ are part of the set, which is just all integers.

We claim that the answer is all $(m,n)$ such that $\boxed{\gcd(m,n) = 1}.$ First, if $\gcd(m,n) = d > 1,$ then obviously every element of $A$ is divisible by $d,$ so we can make $A$ not contain every positive integer. On the other hand, if $\gcd(m,n) = 1,$ by using $m,m$ in the given condition, we see that every multiple of $m^2$ is in $A,$ and similarly every multiple of $n^2$ is also in $A.$ Since $\gcd(m,n) = 1,$ we also have $\gcd(m^2,n^2) = 1,$ so by Bezout we may find integers $k,l$ such that $km^2 + ln^2 = 1.$ Since $km^2, ln^2 \in A,$ we see that $$(km^2)^2 + 2(km^2)(ln^2) + (ln^2)^2 = (km^2+ln^2)^2 = 1 \in A.$$Then taking $1,1$ in the given condition, we see that every integer is in $A.$ Thus all $m,n$ such that $\gcd(m,n) = 1$ work, and these are the only solutions, as claimed.

We claim that the only solutions are coprime pairs. Clearly if they are not coprime, every integer generated will be divisible by their GCD. Now, we show that coprime pairs $(m, n)$ work. Letting $x=y=m$ yields all multiples of $m^2$ are in $A,$ and similarly all multiples of $n^2$ are in $A.$ By Bezout's Theorem, it follows that there are $a, b$ such that $am^2-bn^2=1.$ Thus letting $x=am^2, y=bn^2, k=-2$ yields $(am^2-bn^2)^2 \in A \implies 1 \in A.$ Now letting $x=y=1$ gives $k+2 \in A$ for all integers $k$ and we are done. QED