Take $m = 2n+1 $ for a) and get that all $a$ of the form $4n+1, n \geq 1$ can be achieved.

a) As explained above, if we plug in the pairs $(m, n) = (2k+1, k)$, we can solve to get $a = 4k+1$, so we get $502$ friendly integers. b) This isn't the cleanest solution, but its pretty direct. (There's another solution involving $\upsilon_p$ valuation of both sides.) $2$ is not friendly. So we write $m = kx, n = ky$, so that $gcd(x, y) = 1$. Then plugging these into the equation, we get that $(k^2x^2+ky)(k^2y^2+kx) = 2k^3(x-y)^3 \implies$ \[(kx^2+y)(ky^2+x) = 2k(x-y)^3 (*)\]. Now, we take $\pmod{x}$ to get that $ky^3 \equiv -2ky^3 \implies x \vert 3ky^3 \implies x \vert 3k$ (since $gcd(x, y) = 1$). Similarly, we can get that $y \vert k \vert 3k$. This means that we can write $3k = rxy$ for some integer $r$. Now take $(*)$, multiply both sides by $9$ and substitute $3k = rxy$ to get \[(3kx^2+3y)(3ky^2+3x) = 18k(x-y)^3\]\[(rx^3y+3y)(rxy^3+3x) = 6rxy(x-y)^3\]\[(rx^3+3)(ry^3+3) = 6r(x-y)^3\]. Now taking $\pmod{r}$, we get that $r \vert 9$. So $r = 1, 3, 9$. Now we do a little casework. Case 1: $r = 1.$ This gives $(x^3+3)(y^3+3) = 6(x-y)^3$. If $y \ge 2$, then $(x^3+3)(y^3+3) \ge 6x^3 > 6(x-y)^3$, so no solutions. If $y = 1$, then we can check that we get no solutions (It becomes a cubic with no rational roots). Case 2: $r \ge 3.$ Then, \[(rx^3+3)(ry^3+3) > \left(r(r+3)\right)x^3 \ge 6rx^3 > 6r(x-y)^3\] (since $r(r+3) \ge 6r \Longleftrightarrow r \ge 3$), so no solutions. Since we have exhausted all cases, no solutions.

For b) the direct way is $m=n^2+6n-6+O(n^{-1})$.

Simpler Solution. Notice that $(m^2+n)-(n^2+m)=(m-n)(m+n-1)$, and substitute $y=n^2+m$. then the equation becomes like $y(y+(m-n)(m+n-1))=2(m-n)^3$. So the discriminant of this equation is $D=(m-n)^2((m+n-1)^2+8(m-n))$ is a perfect square, and $(m+n-1)^2+8(m-n)$ is a perfect square, too. Now $(m+n-1)^2+8(m-n)<(m+n+3)^2$, so $(m+n-1)^2+8(m-n)=(m+n+1)^2$. And remaining process is easy^^

a: Pick all $(a, m, n)=(4k+1, 2k+1, k)$. b: Note that $m-n|(m^2+n)-(n^2+m)$. So if $v_p(m^2+n)<v_p(m-n)$ for some $n$, then also $v_p(m+n^2)<v_p(m-n)$. This would imply $v_p([m^2+n][n^2+m])<2v_p(m-n)<v_p(2*[m-n]^3)$, contradiction. So $m-n|n^2+m, m-n|m^2+n$. Let $n^2+m=p(m-n), m^2+n=q(m-n)$ with $pq=2(m-n)$. Then clearly $q>\frac{m^2}{m-n}>m$, which implies $p<\frac{2(m-n)}{m}<2$, so $p=1$. This means $n^2+m=m-n$, which is impossible. So $2$ is not friendly.

I do not know whether it is right but I am posting it.Please point out any flaw and I will be eager to know it. Well if $m$ and $n$ are of different parity then both the expressions $m^2+n$ and $n^2+m$ are odd and so it cannot be a solution.So if $2$ is friendly then then the solutions $m$ and $n$ are of same parity. Now let $m=2^ks_1$ and $n=2^ls_2$ with $k>l$ and $s_1$ and $s_2$ both odd. $(2^{2k}s_1^2+2^ls_2)(2^{2l}{s_2}^2+2^ks_1)=2^{3l+1}(2^{k-l}s_1-s_2)^3$ Powers of $2$ are equal in both sides. if $2l \ge k$ then power in L.H.S is $k+l$.So $k+l=3l+1$.$k=2l+1$ But then $k>2l$. if $2l<k$ then power of $2$ in L.H.S $3l$ which is definitely not equal to the power of $2$ in R.H.S.Remaining case is that of $k=l$ which can be exhausted easily.And the last case if $m$ and $n$ are both odd. The equation can be modified to $mn(n+1)(m-6)+7mn=m^3-3n^3$ If both $m$ and $n$ are odd then R.H.S is even and the first term of L.H.S is also even but the second term is odd.Contradiction!Thus we are done for all cases. Once again I am requesting to point out any flaw.

Oh yeah!And the first part is relatively easy.Well $500$ integers in the set $(1,2,3...2012)$ readily insisted me to think of congruent mpodulo $4$.But I had to really struggle hard to get the required pattern.Strange.So simple but so hard to find.

Here is some motivation behind the solution for part (a): We make the substitution $b := m + n, c:= m - n$ for integers $b, c.$ Therefore $2m = b + c$ and $2n = b - c$, so if we choose $b, c$ of the same parity, $m$ and $n$ are uniquely determined. Meanwhile, we compute \[a = \frac{[(b + c)^2 + 2(b - c)][(b - c)^2 + 2(b + c)]}{16c^3}.\] In particular, $c^3$ must divide the numerator of this fraction, which, after expansion and factorization yields the congruence \[b^2(b + 2)^2 - 2c^2(b^2 - 6b + 2) + c^4 \equiv 0 \pmod{c^3}.\] Because this expression is also divisible by $c^2$, we must have $c \mid b(b + 2).$ At this point, it is natural to let $b$ be a multiple of $c.$ However, after substituting $b = kc$ and chugging through the algebra, we see that this is equivalent to $c^2(4k^2 - 4) \equiv 0 \pmod{c^3}.$ We would like to set $k = 1$, but this results in $b = c$, which is impossible. Instead, we make the substitution $b + 2 = kc$, and the numerator of the above fraction reduces to $4c^2(k^2 - 9) \pmod{c^3}.$ Therefore, we set $k = 3$, which ensures that $b$ and $c$ are of the same parity. We then compute \[a = \frac{b^2(b + 2)^2 - 2c^2(b^2 - 6b + 2) + c^4}{16c^3} = 4c - 3.\] Consequently, we have found the solution family $(a, m, n) = (4c - 3, 2c - 1, c - 1), c \in \mathbb{Z}^{\times}.$ By setting $c = 2, 3, \cdots , 503$, we obtain $502$ solutions for which $m, n \in \mathbb{N}$, as desired.

(b) Let $d = m-n$, $k = m+n-1$. Let $u = n^2+m$. Hence, \[ u(u+kd) = 2d^3 \]. By the quadratic formula, this has solutions only when $\frac{-kd + \sqrt{k^2d^2+8d^3}}{2}$ is an integer. Now $\sqrt{k^2d^2+8d^3}$ must be an integer for any of this to happen, so $\sqrt{k^2+8d}$ is a square as well. We can easily get $\sqrt{k^2+8d} < k+4$, so by parity we get that $\sqrt{k^2+8d} = k+2$. This implies that \[ 4k+4 = 8d \], or $m + n = 2(m-n)$. Hence, if there is a solution, $m = 3n$. It's not too hard from here: \[ (9n^2+n)(n^2+3n) = 16n^3 \], so \[ (9n+1)(n+3) = 16n \]. This implies by the quadratic formula that \[ 9n^2 + 12n + 3 = 0 \], or \[ 3n^2 + 4n + 1 = 0 \], so $n = -1$ or $n = -\frac{1}{3}$. These are not positive, so $2$ is not a friendly number.

For b) For $m=n$ no solutions so wlog $m>n$ $(m^2+n)(n^2+m)=2(m-n)^3$ so $(\dfrac{m^2+n^2+m+n}{m-n})^2+(m+n-1)^2=8(m-n)$ so $8(m-n)>(m-n)^2+(m-n)^2$ so $0<m-n<4$ but obvious from first diophantine $m\pm{n}$ even so $m-n=2$ but then $(m^2+n)(n^2+m)=(n^2+5n+4)(n^2+n+2)=16$ so no solutions...

Here is a new approach for b) based only on divisibility. $(m^2+n)(n^2+m)=2(m-n)^3 \Longleftrightarrow m^3-3n^3=mn(mn+1+6m-6n)$. Trivially we have that $m>n$. Case 1 $3\nmid m,n \implies m\mid n^3$ and $n\mid m^3$ so we have that $\operatorname{rad(n)}=\operatorname{rad(m)}=X$. Let $d=\gcd(m,n) \implies \operatorname{rad(n)}=\operatorname{rad(m)}=\operatorname{rad(d)}=X$. Let $m=da$ and $n=db \implies$ $$m^3-3n^3=mn(mn+1+6m-6n) \Longleftrightarrow d(a^3-3b^3)=ab(d^2ab+1+6da-6db)$$Because $\gcd(d^2ab+1+6da-6db,X)=1$ we have that $X|ab$, but because $\gcd(a,b)=1$ we have that any prime divisor of $X$ is present in the canonical form of either $a$ or $b$ and the prime divisors of both numbers divide $X$ so we must have $\gcd(X,a^3-3b^3)=1$, so $d=ab$. Now a routine check shows that $RHS>LHS$. Other cases can be treated in the exactly same way, after reducing the exponent of $3$ on the both sides.

lol I spent way too long on this problem, I think my solution is overly complicated...

The second troll N4 that I've done today. Solved with L567.

Solved with Alex Zhao, Daniel Yuan, Elliott Liu, Groovy (\help), Isaac Zhu, Raymond Feng, and Ryan Yang Solution 1: For part a, take $m = 2n+1$. Now, we get \[(4n^2 + 5m + 1)(n^2 + 2n + 1) = a(n+1)^3 \Rightarrow a = 4n+1\]Therefore for all $a\equiv 1\mod 4$, we have a solution. This proves that there are at least $500$ friendly numbers between $1$ and $2012$. Now, onto part b. Let $d = \gcd(m,n)$, and let $m = dx, n = dy$. Now, we get \[((dx)^2 + dy)((dy)^2 + dx) = 2(dx-dy)^3 \Rightarrow (dx^{2} + y)(dy^{2} + x) = 2d(x-y)^{3}\]Taking $\mod d$ we get $d | xy$. Taking $\mod y$, we get \[dx^{3} \equiv 2dx^{3}\mod y \Rightarrow y | d\]Taking $\mod x$, we get \[dy^{3}\equiv -2dy^{3}\mod x \Rightarrow x | 3d\]Now, since $d | xy, y | d, x | 3d$, this implies either $d = \frac{xy}{3}$ or $d = xy$. Plugging these two values into our equation gives \[(x^{3} + 1)(y^{3} + 1) = 2(x-y)^{3}\]\[\left(\frac{x^{3}}{3} + 1\right)\left(\frac{y^{3}}{3} + 1\right) = \frac{2}{3}(x-y)^{3}\]In the first case, since $x,y$ are positive, the LHS is greater than $2x^{3}$, while the RHS is less than $2x^{3}$, a contradiction. In the second case, the LHS is greater than $\frac{2}{3}x^{3}$, while the RHS is less than $\frac{2}{3}x^{3}$, a contradiction. Since both cases give a contradiction, this means that $a=2$ is not friendly. Solution 2: For the first part, reference the above solution For the second part, we claim that there are no solutions. Suppose FSoC that $(m^2+n)(m+n^2)=2(m-n)^3$ holds for some pair of positive integers $m$ and $n$. Trivially, we must have $m>n$. Then we have that \[m^3-m^2(n^2+6n)+m(6n^2-n)-3n^3=0\]Define $f(m)=m^3-m^2(n^2+6n)+m(6n^2-n)-3n^3$ (treat $n$ as fixed). Note that \[m\leq n^2\implies f(m)\leq -6m^2n+6mn^2-mn-3n^3<0,\]so we must have $m>n^2$. Also note that \[m\geq n^2+6n\implies f(m)\geq (n^2+6n)(6n^2-n)-3n^3=(n^3+6n^2)(6n-1)-3n^3>0.\]Thus, $m<n^2+6n$. Furthermore, we have that $f'(m)=3m^2-2(n^2+6n)m+6n^2-n$ which has a larger root that is at most $\frac{2(n^2+6n)}3$. We set aside $n<12$ as a finite case check. For $n\geq 12$, we have that the largest root of $f'(m)$ is at most $n^2$, so $f(m)$ is increasing on $(n^2,n^2+6n)$. We now have that \[f(n^2+6n-6)=(n^2+6n-6)(36-37n)-3n^3<0,\]so $m=n^2+6n-i$ for some $1\leq i\leq 5$. This gives (by taking the original equation mod $n$) that $m^3\equiv0\pmod n$, so $n\mid i^3$ for some $1\leq i\leq 5$. Combined with the cases of $n<12$, this yields the following values of $n$: \[n\in\left\{ 1,2,3,4,5,6,7,8,9,10,11,27,16,32,64,25,125 \right\}.\] Checking that these $n$ give no solutions is easy and is left as an exercise to the reader.

Eyed wrote: Solved with...Groovy (\help) Who is this Groovy??

IMO shortlist 2012 N4

Simple: Let $x=m-n$. The equation rewrites as $(n^2+(2x+1)n+x^2)(n^2+n+x) =ax^3$. Clearly $x|n^2+n$ so let $n^2+n=kx$. The equation becomes $(k+1)(k+2n)=x(a-k-1)$ For part (a), it is extremely convenient to put $x=k+1$ giving $m=2n+1$ and $a=4n+1$. For part (b), note that $k \geq 1$ so the last equation has RHS $ \leq 0$ but the LHS is positive, CONTRADICTION. So $2$ is not friendly.

Part A Claim: All $a \equiv 1 ( \mod 4)$ works for all $m=2n+1$ Proof: $$(4n^2+5n+1)(n+1)^2=a(n+1)^3$$$$4n^2+n(5-a)+1-a=0$$ $$n=\frac{a-5 \pm \sqrt{a^2+6a+9}}{8}$$$$=\frac {a-5 \pm a+3}{8}$$$$= 1 or \frac{a-1}{4}$$ Therefore, we clearly have $500$ such working $n$. Part B Subclaim: If $p|m$, then $p|n$ also,$p|n$, then $p|=m$ for $p$ not equal to $3$ Claim: $V_p(m)=2V_p(n)$ or $V_p(n)=2V_p(m)$ for all prime $p$, where $p$ divides n Proof: $$(m^2+n)(n^2+m)=2(m-n)^3$$$$0=m^3+m^2(-6n-n^2)+m(6n^2-n) -3n^3$$ Which implies that $m|3n^3$ and $n|m^3$ So, our subclaim is true. Now, consider some prime $p$ such that $p|n$ which also implies that $p|m$ Case 1: $V_p(m)> V_p(n)$ $$V_p(m^2+n)=V_p(n)$$$$V_p2(m-n)^3 \geq 3V_p(n)$$ Which implies that $V_p(n^2+m) \geq 2V_p(n)$ As such, $V_p(m)=V_p(n^2)=2V_p(n)$ Case 2: $V_p(n)= V_p(m)$ $$V_p(m^2+n)=V_p(n)$$$$V_p(n^2+m)=V_p(n)$$$$V_p(2(m-n)^3) \geq 3V_p(n) >2V_p(n) $$ Which is absurd, hence such a case cannot exist to begin with. Therefore, our main claim is true Case 1: $3$ does not divide $m$ Observe that this directly implies that $m|n^2$ $$2m^3>2(m-n)^3 =(m^2+n)(n^2+m) >2m^3$$ Which is absurd Case 2: $3$ divides $m$ If $m=3n^2$ $$(9n^4+n)(10n^2)=2(3n^2-n)^3$$$$90n^3+10=2(27n^3-27n^2+27n-1)$$$$ \leq 2(27n^3)=54n^3$$ But this RHS is smaller than the LHS which is absurd, implying that there is no solution If $m=\frac{3n^2}{2}$ $$(\frac{9n^2}{4}+n)(\frac{5n^2}{2}=2(\frac{3n^2}{2}-n)^3$$$$45n^3+20=54n^3-108n^2+72-16$$$$0=9n^3-108n^2+72n-36$$ Checking from $n=1$ to $n=12$, we observe that there is no $n$ satisfying the above equation Note that for the remaining cases $m\leq n^2$, which as we have resolved earlier fails due to size argument Therefore $a=2$ is not friendly

Solved with WLOGQED1729 who think it's N1 level ORZ ORZ. a) Take $(m,n)=(2n+1,n)$ gives $a=4n+1$. So these $500$ numbers $1,5,9,\dots,1997\in\{1,2,\dots, 2012\}$ are friendly integers. b) The answer is negative. Obviously, $m>n$. We do some algebraic manipulation to get the following: $$(m+n^2)^2+(m+n^2)((m-n)(m+n-1))-2(m-n)^3=0.$$Consider the discriminant of the quadratic equation $m+n^2$. We have $$(m-n)^2(m+n-1)^2+8(m-n)^3\in\mathbb{Z}^2\implies (m+n-1)^2+8(m-n)\in\mathbb{Z}^2.$$But note that $(m+n-1)^2<(m+n-1)^2+8(m-n)<(m+n+3)^2$. Thus, $$(m+n-1)^2+8(m-n)\in\{(m+n)^2,(m+n+1)^2,(m+n+2)^2\}.$$Moreover, due to parity reasoning, $(m+n-1)^2+8(m-n)\equiv (m+n+1)^2$. Hence, $(m+n-1)^2+8(m-n)=(m+n+1)^2$. This gives $m=3n$. But we have a contradiction since $$((3n)^2+n)(n^2+3n)=2(3n-n)^3\implies 9n^2(3n+1)(n+1)=0$$has no solution in positive integers, and we're done.

Claim $a=4l+1$ is friendly integer for all integers $l \in \{1,2,...,500\}$. We choose $(m,n)=(2l+1,l)$ and found that \begin{align*} \left(m^2+n\right)\left(n^2+m\right)&=\left((2l+1)^2+l\right)\left(l^2+2l+1\right)\\ &=\left(4l^2+5l+1\right)\left(l^2+2l+1\right) \\ &= (4l+1)(l+1)^3\\ &=a(m-n)^3 . \end{align*}Hence, there are at least $500$ friendly integers in the set $\{ 1,2,\ldots ,2012\}$. Claim $a=2$ isn't a friendly integer. Considering $m >n$ \begin{align*} 0&=\left(m^2+n\right)\left(n^2+m\right)-2(m-n)^3 \\ &= m^2n^2+6m^2n+3n^3+mn-m^3-6mn^2 \\ &> m^2n^2+3n^3+mn-m^3 \\ &> m^2n^2-m^3 \Rightarrow m>n^2 \end{align*}First, let $m=dm_1,n=dn_1$ ; $gcd(m_1,n_1)=1$. Considering \[\left(m^2+n\right)\left(n^2+m\right)-2(m-n)^3=m^3-6m^2n+6mn^2-m^2n^2-mn-3n^3\]and found that \begin{align*} 0&=dm_1^3-6dm_1^2n_1+6dm_1n_1^2-3dn_1^3-d^2m_1^2n_1^2-m_1n_1 \\ &=m_1(dm_1^2-6dm_1n_1+6dn_1^2-d^2m_1n_1^2-n_1)-3dn_1^3. \end{align*}Since $gcd(m_1,n_1)=1$ implies that \[ m_1 \ | \ 3dn_1^3 \Rightarrow m_1\ | \ 3d \Rightarrow dm_1 \ | \ 3d^2 \ | \ 3n^2 \Rightarrow m \ | \ 3n^2 . \]Therefore $m=3n^2,\frac{3n^2}{2}$. Proof is left as an exercise to the reader.

Part A: Set $m = 2n+1$. We see that $a = 4n + 1$ is friendly for any positive integer $n$. Since $a\in \{1,2,\ldots, 2012\}$ for $1\le n\le 500$, we have at least $500$ friendly integers. Part B: No, $a=2$ is not friendly. Suppose it was. Then there exist positive integers $m,n$ satisfying \[(m^2 + n)(n^2 + m) = 2(m-n)^3\] Claim: $m-n \mid m^2 + n$. Proof: Suppose not. Then there exists $p$ prime where $\nu_p(m-n) > \nu_p(m^2 + n)$. Let $t = \nu_p(m^2 + n)$. Since the difference between $m^2 +n$ and $n^2 + m$ is a multiple of $p^{t+1}$ (since it's a multiple of $m-n)$, we get that $\nu_p(n^2 + m) = t$ also. Hence the $\nu_p$ of the LHS is $2t$ and the the $\nu_p$ of the RHS is at least $3\nu_p(m-n)$, which exceeds $3t>2t$, absurd. $\square$ Let $m^2 + n = k(m-n)$, where $k$ is an integer. Since $(m^2 + n)(n^2 + m)$ is positive, $2(m-n)^3$ is also positive, so $m-n$ is positive. Thus, $k$ is a positive integer. We have $n^2 + m = (k - (m + n - 1))(m-n)$. The equation simplifies to $k(k - (m + n - 1)) = 2(m-n)$, so \[ k^2 - (m + n - 1)k - 2(m-n) = 0\]The discriminant of this equation must be a perfect square since $k$ is an integer. Thus, $(m+n-1)^2 + 8(m-n) = d^2$ for some positive integer $d$. Clearly $d$ is the same parity of $m + n - 1$, and $d> m + n - 1$. If $d\ge m + n + 3$, then \begin{align*} d^2 - (m + n - 1)^2 \ge (m + n + 3)^2 - (m + n - 1)^2 \\ = 8m + 8n + 8 > 8(m-n), \\ \end{align*}absurd. Therefore $d = m+ n + 1$. Then $8(m-n) = d^2 - (m + n - 1)^2 = 4m + 4n$, so $4m = 12n\implies m = 3n$. Plugging gives $(9n^2 + n)(n^2 + 3n) = 16n^3$. Dividing both sides by $n^2$ gives $(9n + 1)(n+3) = 16n$, so $9n^2 + 12n + 3 = 0\implies 3n^2 + 4n + 1 = 0$. This can be factored as $(3n + 1)(n + 1)$, so $n = -1$ or $n = -1/3$, which is absurd since $n$ is positive. Therefore $2$ is not friendly.

Let $m=2n+1$ then $a=4n+1$ works, so all $5,9,13,\dots,2009$ work, and there are at least $500$ of them. We claim that $2$ is not friendly. Since $m^2+n\equiv n^2+m\pmod {m-n}$, if $m-n\nmid m^2+n$ then $m-n\nmid n^2+m$ so $(m-n)^3\nmid (m^2+n)(n^2+m)$. Thus, let $m^2+n=x(m-n)$ and $n^2+m=y(m-n)$. We have $xy(m-n)^2=2(m-n)^3$ so $xy=2(m-n)$. We have $y=\tfrac{2(m-n)^2}{(m^2+n)}<2$ so $n^2+m=m-n$ implying $n=1$. Then, $(m^2+1)(m+1)=2(m-1)^3$. We get \begin{align*} 2m^3-6m^2+6m-2 &= m^3+m^2+m+1 \\ m^3-7m^2+5m-3 &= 0 \end{align*}Clearly, there are no solutions for $m$.

For part a, simply take $m = 2n+1$ to finish. For part b, simply note that if $v_p(m-n) > v_p(n^2+n)$ then expanding $v_p$ gives bad news. Hence $m-n \mid n^2+n$ but then by bounding, we have that $a = (1+\frac{2n}{m-n}+\frac{n^2+n}{(m-n)^2})(1+\frac{n^2+n}{m-n})$ but the latter term is at least $2$ and the first term is strictly greater than $1$, so $a = 2$ is impossible.