Right it’s only (0,0), should’ve checked properly... nevertheless, bounding after a<0,b<0 should work... Hmmm... only a -2 for that (0,-n^6) error—a pleasant surprise indeed!

ubermensch wrote: $(0,-n^6)$ and $(-n^6,0)$?... bounding a, b seemed to work after getting a,b<0 That doesn't work... All the roots need to be integers, not just one. I got only solution as (0,0)

Haha, someone else also fell for $(0, -n^6)$ A tweak of the official solution:

I forgot about (0,0) and wrote no solutions. Can I expect 15?

Nice problem

Aarth wrote: I forgot about (0,0) and wrote no solutions. Can I expect 15? you missed a trivial solution, idk how much marks you will get.

How many marks will I lose if Ididnt verify the ans ? (as in didnt write that $x^3=0$ has integral roots)

This was a cool infinite decent. Just show that there are no nonzero integers $p,q,r,s$ such that $$pq(p+q) \mid r^2 + rs + s^2, rs(r+s) \mid p^2 + pq + q^2$$ That's it. This completes the proof.

Any solution using elliptic curves?

Redacted

You can't subtract : )

DapperPeppermint wrote: anantmudgal09 wrote: Find all pairs of integers $(a,b)$ so that each of the two cubic polynomials $$x^3+ax+b \, \, \text{and} \, \, x^3+bx+a$$has all the roots to be integers. Proposed by Prithwijit De and Sutanay Bhattacharya Subtract to get $(a-b) x=a-b$. If a=b using vieta find (a+1) (b+1)(c+1)=1, at least one of these is 0 due to sign of product being positive so a =b=0. If a≠b and x=1(both can't hold true), then every root of both the equations must satisfy (a-b) x=a-b which can't be possible if root is not 1 and a-b is not 0 so all roots are 1 which gives contradiction. Wait that doesn't work... U can't just assume they have a common root right?

I actually thought in the exam that (0,0) is a little strange, but wrote it anyways... well at least one problem down...

N1RAV wrote: You can't subtract : ) I know I'm dumb and read in haste IMOTC hopes gone

My solution was quite different but got the same answer. I took the roots as $x,y,z$ and $p,q,r$. $(x+1)(y+1)(z+1)=a-b+1, (p+1)(q+1)(r+1)=b-a+1$ If $a=b$, it's obvious they are zero Else, WLOG $a>b\implies (p+1)(q+1)(r+1)\leq 0$. Rest was easy casework

Looks like my solution's different from all of the above, so ig i'll post mine too.. First consider the case when one of $a,b$ is $0$ and then easily get that this means that $a = b = 0$ Replace $a,b$ with $-y,-z$ so that $y,z$ are positive. Let the roots of the first equation be $-a_1, -a_2, a_1 + a_2$ and let the roots of second equation be $-b_1, -b_2, b_1 + b_2$ so that all these variables are positive too From vieta's we get that $a_1^2 + a_2^2 + a_1a_2 = y, a_1a_2(a_1+a_2) = z$ and $b_1^2 + b_2^2 + b_1b_2 = z, b_1b_2(b_1+b_2) = y$ If one of $a_1, a_2, b_1, b_2$ is $1$, WLOG $a_1 = 1$, but this means $y+z = 1$ and so $b_1 = 1$ as well. But then substituting gives $a_2 = b_2 = -1$, which is a contradiction since they are supposed to be positive. So, $a_1, a_2, b_1, b_2 \ge 2$. WLOG $a_1 \ge a_2$. Observe that the following inequality holds: $a_1^2(a_2 - 1) + a_2(a_1a_2 - a_1 - 1) > 0$ since $a_2 > 1$ and $a_1a_2 \ge 2a_1 = a_1 + a_1 \ge a_1 + 2 > a_1 + 1$ which rearranges to $a_1a_2(a_1 + a_2) > a_1^2 + a_2^2 + a_1a_2$, which means $y > z$. Similarly, we can do the same thing for $b_1, b_2$ which gives $z > y$, which is a contradiction. Therefore, the only solutions to this equation are $(a,b) = (0,0)$

Math-wiz wrote: My solution was quite different but got the same answer. I took the roots as $x,y,z$ and $p,q,r$. $(x+1)(y+1)(z+1)=a-b+1, (p+1)(q+1)(r+1)=b-a+1$ If $a=b$, it's obvious they are zero Else, WLOG $a>b\implies (p+1)(q+1)(r+1)\leq 0$. Rest was easy casework That's almost how I did it!

Looks different from all of the above If one of $a,b$ is $0$, then it's not tricky to see that the other is $0$ too. We now ignore this case. By differentiating, we can see that $a < 0$ and $b < 0$ Let the roots of $P(x) = x^3 + ax + b$ be $p\le q \le r$ and, $Q(x) = x^3 + bx + a$ be $u \le v \le w$ It's not hard to see that exactly two roots of each polynomial are negative. We now bound. WLOG $|a| \ge |b|$ $|q^3| \le \frac{|qp(q+p)|}{2} \le \frac{b}{2}$, so $0 = q^3 + aq + b \ge \frac{3b}{2} + aq$, so $|3b| \ge |2aq|$. Since $|a| > |b|$, we have that $q = -1$ and plugging that in, $-1 -a + b = 0$ By using the exact same bounds with $v$ instead of $q$, $|3a| \ge |2vb|$ and since $v \neq -1$, $|3a| \ge |4b|$. We can easily test out the values of $(a,b) = (-4,-3), (-3,-2), (-2,-1)$ by hand and none of these work. Thus there are no solutions for which $ab \neq 0$

Let , $$ P_1(x)=x^3+ax+b \, \, \text{and} \, \,P_2(x)= x^3+bx+a$$ Let, $ b_1, b_2,b_3$ are roots of $P_1$ and $a_1,a_2,a_3$ are roots of $P_2$. so, \[b_1+b_2+b_3 =0 , b_1b_2+b_2b_3+b_3b_1 =a \implies b_1^2 +b_2^2 +b-3^2 =-2a\] clearly atleast one of $b_i$ is divisable by 2 . so , $2\mid b_1b_2b_3 = -b $ , $2 | b$ . similarly , $ 2 | a $ now ,\[2 \mid 0= b_i^3 +ab_i +b \implies 2\mid b_i^3 \implies 8 \mid b_i \] so, $ 8^3 \mid b$ and for $a $ same . using same step get $8^6 \mid a,b$ using infinite descent get , only possibility is $a,b=0$ whoever thinks about , $a,b = 0,-n^6$ forgot about complex numbers .

Denote by $x_1,x_2,x_3$ the roots of $x^3+ax+b=0$ and by $y_1,y_2,y_3$ the roots of $x^3+bx+a=0$. As for $f(x)=x^3+ax+b$, $f'(x)=3x^2+a>0$ for $a>0$ (unless we have 3 repeated roots, which forces $a=b=0$ anyway), we must have $a \leq 0$. Similarly, $b \leq 0$. Also, WLOG, assume $|a| \geq |b|$. Case 1: Neither of $a,b$ is 0 (which forces all of the roots to be non-zero) As $x_1+x_2+x_3=0$ and $x_1x_2x_3=-b>0$, we must have 2 of $x_1,x_2,x_3$ to be negative and 1 to be positive. WLOG say $x_3=-(x_1+x_2)>0$. $x_1x_2+x_2x_3+x_3x_1 = x_1x_2-(x_1+x_2)^2=-(x_1^2+x_2^2+x_1x_2)=a \implies x_1^2+x_2^2+x_1x_2=-a \implies |a|=x_1^2 +x_2^2+x_1x_2$. $x_1^2x_2+x_2^2x_1=b \implies |b|=-(x_1^2x_2+x_2^2x_1)$. Letting $m=-x_1,n=-x_2$ such that $m,n>0$, we get $m^2n+mn^2 \leq m^2+n^2+mn$. Say $m \geq n$. Clearly, $mn^2 \geq n^2 \implies m^2(n-1) \leq n^2 \implies n-1 \leq 0 \implies n \leq 1 \implies n=1$. Thus $-1$ is a root of $x^3+ax+b=0 \implies b=a+1$. Now we work on $x^3+bx+a=0$. Again, as $y_1+y_2+y_3=0$ and $y_1y_2y_3=-a>0$, WLOG we let $y_3=-(y_1+y_2)>0$ and $y_1,y_2<0$. $y_1y_2-(y_1+y_2)^2=-(y_1^2+y_2^2+y_1y_2)=b=a+1$, $y_1^2y_2+y_2^2y_1=-a \implies -(y_1^2+y_2^2+y_1y_2+1)=y_1^2y_2+y_2y_1^2$. Letting $m=-y_1, n=-y_2$, and WLOG assuming $m \geq n >0$, we get $m^2+n^2+mn+1=m^2n+mn^2$. Say $n \geq 2$. Clearly we have $mn^2>n^2+1$ and $m^2n \geq 2m^2 \geq m^2+mn \implies$ Contradiction. This forces $n=1 \implies -1$ is a root of $x^3+bx+a=0 \implies a=b+1$. We now have forced both $a=b+1$ and $b=a+1$, and can thus conclude that this case yields no solution. Case 2: At least one of $a,b$ is $0$ WLOG say $a=0$. Notice (as I should have certainly noticed during INMO) that $x^3+b=0$ has at least 2 complex roots for $b$ non-zero, hence forcing $a=b=0$, which clearly works. Thus the only solution is $(a,b)=(0,0)$.

I have proved that $a$ or $b$ cant be odd with help of that i have proved they both cant be 2mod(4) and like that for general, that it can't be $2^k mod(2^{k+1})$ if $a,b\equiv 0 mod(2^k)$ so must be divisible by $2^n \forall n\epsilon R$ so they both should be zero. I have done this by using $$-a=\alpha\beta\gamma$$$$b=\sum \alpha\beta$$$$\alpha+\beta+\gamma=0$$and taking the respective mods and making few cases.

The following is an evil solution I found in contest- Case 1: $ab=0$ WLOG $a=0$, now $b=0$ and $(0,0)$ works. Now, assume that $ab\ne 0$. So, we have that the $6$ integer roots are all non-zero. Now, WLOG $|a|\ge |b|$. Now, let $\alpha, \beta,\gamma$ be roots of $x^3+ax+b$. Thus, $|\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}|=\frac{a}{b}$. WLOG, $|\frac{1}{\alpha}|\ge |\frac{1}{\beta}|,|\frac{1}{\gamma}|$. Thus, $|\frac{1}{\alpha}|\ge \frac{1}{3}\implies |\alpha|\le 3$. Now, we do casework Case 1: $\alpha=1$, we have $b=-a-1$ and $x^3+ax-a-1=(x-1)(x^2+x+a+1)$ and $x^3+(a+1)x+a=(x-1)(x^2-x-a)$ have integer roots but then $\sqrt{1+4(1+a)}$ and $\sqrt{1-4a}$ are both integers which is absurd.\ Case 2: $\alpha=-1$, we have $b=a+1$. Thus, $x^3+(a+1)x+a=(x-\alpha')(x-\beta')(x-\gamma')$ for integer roots $\alpha, \beta, \gamma$ of $x^3+bx+a$, thus $-2=(-1-\alpha')(-1-\beta')(-1-\gamma')$ but as none of the roots are $0$, we must have that two are $-2$ and one is $1$ but these do not sum to $0$. Case 3: $\alpha=2$, we have $b=-2a-8$ but $|a|\ge |b|=|2a+8|$ hence $|a|\le 8$ and thus $|b|\le 8$ but since $\alpha=2$ all roots have mod atleast $2$ of $x^3+ax+b$ thus if $|b|=8$ then all roots must be $\pm 2$ but these cannot sum to $0$. Case 4: $\alpha=-2$, same as case $3$ as $b=2a+8$ Case 5: $\alpha=\pm 3$, as $|\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}|=\frac{a}{b}$ we get that all roots are $\pm 3$ but these cannot sum to $0$. Thus, we are done with all cases!

My solution is of more algebraic flavour I feel. anantmudgal09 wrote: Find all pairs of integers $(a,b)$ so that each of the two cubic polynomials $$x^3+ax+b \, \, \text{and} \, \, x^3+bx+a$$has all the roots to be integers. Proposed by Prithwijit De and Sutanay Bhattacharya Solution - First note that if $P(x)$ has at least two reals roots then $P'(x)$ has at least one , this gives $a,b \leq 0$ Now if any one of $a,b$ is zero so is the other and we get the solution $(0,0)$ , so assumed $a,b<0$ Consider $x^3+ax+b=0$, product of roots is $-b >0$ and sum is zero , hence the roots are two negative and one positive. So let them be $-p, -q , p+q$ where $p,q$ are positive integers ($0$ can't be root at $a,b$ are non zero) And by Vietta's $-b = pq(p+q)$ and $-a = p^2+q^2+pq$ So we have $S_{ab} = (-b)-(-a) = pq(p+q) - p^2 - q^2 - pq$ And after doing exact same for other equation with roots $-r,-s,r+s$ we have $S_{ba}=(-a)-(-b) = rs(r+s)-r^2-s^2-rs$ Now $S_{ab} = -S_{ba}$ so at least one of them must be non-positive, say $S_{ab}\leq 0$ $\implies pq(p+q) - p^2 - q^2 - pq \leq 0 \implies p^2(q-1)+q^2(p-1)-pq \leq 0$ Now if $p,q \geq 2 , p^2(q-1)+q^2(p-1)-pq \geq p^2+q^2-pq > 0$ a contradiction. So assume $p=1$ , this gives $S_{ab} = -1$ and hence $S_{ba} = 1$ Hence $rs(r+s)-r^2-s^2-rs = 1$ , now if $r$ or $s$ is $1$ then this is $-1$ , hence we have $r,s \geq 2$ So $1 = rs(r+s)-r^2-s^2-rs = r^2(s-1)+s^2(r-1)-rs \geq r^2+s^2 -rs \geq (r-s)^2 + rs \geq rs > 1$ A contradiction !! Hence $(a,b)=(0,0)$ is the only solution.

It is easy to see (for instance by taking derivative) that $a,b \leq 0$. So let the roots be $-p,-q,p+q$ and $-r,-s,r+s$ respectively where $p,q,r,s \geq 0$. Then $$-a=rs(r+s)=p^2+pq+q^2$$$$-b=pq(p+q)=r^2+rs+s^2$$WLOG $s= \min \{p,q,r,s \}$. Case 1: $s=0$ In this case we get $p^2+pq+q^2=0$ $\implies$ $p=q=0$ $\implies$ $a=b=0$. Case 2: $s=1$ In this case we get $$pq(p+q)=r^2+r+1=p^2+pq+q^2+1$$$$\implies pq(p+q+1)=(p+q)^2+1$$$$\implies p+q+1 \mid 2$$$$\implies p+q \leq 1$$Contradiction since $p,q \geq 1$. Case 3: $s \geq 2$ In this case we have $rs \geq r+s$ and $pq \geq p+q$. So $$p^2+pq+q^2 \geq (r+s)^2 > r^2+rs+s^2$$and $$r^2+rs+s^2 \geq (p+q)^2 > p^2+pq+q^2$$Contradiction! So the only solution is $$\boxed{a=b=0}$$

Is there any strong boy/girl who tried to use $\textbf{Cardan's method}$ or the $\textbf{Cubic Discriminant}$ for this problem? I thought that for a moment and gave up in 1 minute straight after seeing the horrible looking calculation...

anantmudgal09 wrote: Find all pairs of integers $(a,b)$ so that each of the two cubic polynomials $$x^3+ax+b \, \, \text{and} \, \, x^3+bx+a$$has all the roots to be integers. Proposed by Prithwijit De and Sutanay Bhattacharya Seems correct We can more generally work in $\mathbb{R}$ for ideas and themes and then transition to the case of $\mathbb{Z}$ with some workwhich ultimately takes us to the solution.

Therefore for $p(x)=x^3 +ax+b$ and $q(x)=x^3 +bx+a$ to have real roots we must have $$\begin{aligned}27a^2 +4b^3 \le 0 \\ 27b^2 +4a^3 \le 0 \end{aligned}$$Let's define $\sigma_j$ as elementary symmetric polynomials in one variable(technically speaking we associate them to a polynomial in $\mathbb{R}[x]$ with $a,b$ as roots) , then adding the inequalities we get $27(\sigma_{1}^2 -2\sigma_2)+4(\sigma_{1}^{3}-3\sigma_1 \sigma_2 )=\mathcal{L} \le 0$ As equality occurs at $a=b=0$ we assume $a,b \neq 0$ Then using Maclaurin's inequalities we get $$\frac{\sigma_1}{2} \ge \sqrt{\frac{\sigma_2}{1}} \Rightarrow \sigma_2 \le \frac{\sigma_{1}^2}{4}$$Using this we get $\mathcal{L} \ge \frac{27\sigma_1^{2}}{2} +\sigma_1^{3}$ ,.Then we note that as $p,q$ have real zeros $p^{'} ,q^{'}$ have at least one real zero, this forces $a,b \le 0 \Leftrightarrow \sigma_1 \le 0$ Therefore we get $\mathcal{L} \ge \sigma_1^{2} \left(\frac{27}{2}+\sigma_1 \right) $ then we note setting $\sigma_1 =\frac{-27}{2}$ leads to solutions in $\mathbb{Q}$, if $\sigma_1 \ge \frac{-27}{2}$ we've a contradiction and if we set $\sigma_1 \le \frac{-27}{2}$ then we can work as follows : We can assume without loss of generality $a > b $ then from the individual inequalities we're forced to have $a < \frac{-27}{4}=\frac{a+b}{2} $ , which is a contradiction, Therefore we can conclude that the cubics $p,q$ are irreducible in $\mathbb{Z}[t]$ for all the case of $a,b \neq 0$ Therefore we can conclude the only solution to be $(a,b)=(0,0)$

homotopygroup wrote: anantmudgal09 wrote: Find all pairs of integers $(a,b)$ so that each of the two cubic polynomials $$x^3+ax+b \, \, \text{and} \, \, x^3+bx+a$$has all the roots to be integers. Proposed by Prithwijit De and Sutanay Bhattacharya Looks correct. We can more generally work in $\mathbb{R}$, which ultimately takes us to the solution.

Therefore for $p(x)=x^3 +ax+b$ and $q(x)=x^3 +bx+a$ to have real roots we must have $$\begin{aligned}27a^2 +4b^2 \le 0 \\ 27b^2 +4a^3 \le 0 \end{aligned}$$Let's define $\sigma_j$ as elementary symmetric polynomials in one variable(technically speaking we associate them to a polynomial in $\mathbb{R}[x]$ with $a,b$ as roots) , then adding the inequalities we get $27(\sigma_{1}^2 -2\sigma_2)+4(\sigma_{1}^{3}-3\sigma_1 \sigma_2 )=\mathcal{L} \le 0$ As equality occurs at $a=b=0$ we assume $a,b \neq 0$ Then using Maclaurin's inequalities we get $$\frac{\sigma_1}{2} \ge \sqrt{\frac{\sigma_2}{1}} \Rightarrow \sigma_2 \le \frac{\sigma_{1}^2}{4}$$Using this we get $\mathcal{L} \ge \frac{27\sigma_1^{2}}{2} +\sigma_1^{3}$ ,.Then we note that as $p,q$ have real zeros $p^{'} ,q^{'}$ have at least one real zero, this forces $a,b \le 0 \Leftrightarrow \sigma_1 \le 0$ Therefore we get $\mathcal{L} \ge \sigma_1^{2} \left(\frac{27}{2}+\sigma_1 \right) >\frac{27\sigma_1^{2}}{2} >0$, which is a contradiction Therefore we can conclude the only solution to be $(a,b)=(0,0)$

Wait a minute, was this a cubic discriminant?

homotopygroup wrote: anantmudgal09 wrote: Find all pairs of integers $(a,b)$ so that each of the two cubic polynomials $$x^3+ax+b \, \, \text{and} \, \, x^3+bx+a$$has all the roots to be integers. Proposed by Prithwijit De and Sutanay Bhattacharya Looks correct. We can more generally work in $\mathbb{R}$, which ultimately takes us to the solution.

Therefore for $p(x)=x^3 +ax+b$ and $q(x)=x^3 +bx+a$ to have real roots we must have $$\begin{aligned}27a^2 +4b^2 \le 0 \\ 27b^2 +4a^3 \le 0 \end{aligned}$$Let's define $\sigma_j$ as elementary symmetric polynomials in one variable(technically speaking we associate them to a polynomial in $\mathbb{R}[x]$ with $a,b$ as roots) , then adding the inequalities we get $27(\sigma_{1}^2 -2\sigma_2)+4(\sigma_{1}^{3}-3\sigma_1 \sigma_2 )=\mathcal{L} \le 0$ As equality occurs at $a=b=0$ we assume $a,b \neq 0$ Then using Maclaurin's inequalities we get $$\frac{\sigma_1}{2} \ge \sqrt{\frac{\sigma_2}{1}} \Rightarrow \sigma_2 \le \frac{\sigma_{1}^2}{4}$$Using this we get $\mathcal{L} \ge \frac{27\sigma_1^{2}}{2} +\sigma_1^{3}$ ,.Then we note that as $p,q$ have real zeros $p^{'} ,q^{'}$ have at least one real zero, this forces $a,b \le 0 \Leftrightarrow \sigma_1 \le 0$ Therefore we get $\mathcal{L} \ge \sigma_1^{2} \left(\frac{27}{2}+\sigma_1 \right) >\frac{27\sigma_1^{2}}{2} >0$, which is a contradiction Therefore we can conclude the only solution to be $(a,b)=(0,0)$

This solution seems false. Already in $\mathbb{Q}$ we have additional pairs other than $(0,0)$, for example $a=b=-\frac{27}{4}$, where $x^3-\frac{27x}{4}-\frac{27}{4}$ has roots $-\frac{3}{2},-\frac{3}{2},3$.

@above, I see, I missed to specifically make the transition from results in $\mathbb{R}$ to conclusion in $\mathbb{Z}$ I've made some edits, also please point out, if there's some more mess I've done

anantmudgal09 wrote: Find all pairs of integers $(a,b)$ so that each of the two cubic polynomials $$x^3+ax+b \, \, \text{and} \, \, x^3+bx+a$$has all the roots to be integers. Proposed by Prithwijit De and Sutanay Bhattacharya Let's assume the integral roots for $x^3+ax+b=0$ and $x^3+bx+c=0$ are $\{a_1, a_2, a_3\}$ and $\{b_1, b_2, b_3\}$ respectively.. Now we have $a_1+a_2+a_3=0,b_1+b_2+b_3=0$ and all $a_i,b_i\in Z$ So clearly only possibility we have is either all roots are even or any $1$ must be even with $2$ odd. So Case 1-: when all roots $(a_1,a_2,a_3)$ are even then Claim 1-: as $a_1+a_2+a_3=0$ So exactly for $2$ roots (say a_1,a_2) we must have $v_2(a_1)=v_2(a_2)$ with $v_2(a_3)>v_2(a_1)$ Proof-: FTSOC assume $v_2(a_1)>v_2(a_2)>v_2(a_3)$ and $2^{v_2(a_i)}*k_i=a_i$ for $i=1,2,3$ then we will have $k_3+2^{v_2(a_2)-v_2(a_3)}*k_2+2^{v_2(a_1)-v_2(a_2)}*k_3=0$ which is clearly not possible as sum of $2$ even and $1$ odd term can never be $0$. Hence we must have $v_2(a_1)=v_2(a_2)$ with $v_2(a_1)<v_2(a_3)$ to have $2$ odd terms and a even term. Now as $a_1*a_2+a_2*a_3+a_3*a_1=a$ So ${v_2(a_1)}=v_2(a)$ and $a_1*a_2*a_3=b\implies 2*v_2(a_1)+v_2(a_3) =v_2(b)\implies v_2(b)>v_2(a)$ And by second equation we also have $b_1*b_2*b_3=-a,b_1*b_2+b_1*b_3+b_3*b_2=b$ Then Clearly all $b_1,b_2,b_2$ must also be even By claim 1 $v_2(b_1)=v_2(b_2)$ with $v_2(b_1)<v_2(b_3)$ and we also get $2*v_2(b_1)+v_2(b_3)=v_2(a_1)$ and $2*v_2(b_1)=2*v_2(a_1)+v_2(a_3)$ which is clearly Not possible So $a_1,a_2,a_3$ can't be even. Case 2-: When only one root is even (say $a_1$) and other $2$ $(a_2,a_3)$are odd Then $a_1*a_2+a_2*a_1+a_3*a_2=a\implies a$ is odd $\implies b_1*b_2*b_3=-a\implies$ all root of second equation is odd which is not possible. So no solution is there other then trivial solution which is obviously $(a, b) =(0, 0)$ $\blacksquare$

anantmudgal09 wrote: Haha, someone else also fell for $(0, -n^6)$ A tweak of the official solution:

Could anyone please tell me briefly what is denoted by $\mu_2(\alpha)$, $\mu_2(\beta)$? Never seen such notation before in this context.

rafaello wrote: Could anyone please tell me briefly what is denoted by $\mu_2(\alpha)$, $\mu_2(\beta)$? Never seen such notation before in this context. It's just $v_2(n)$, which is the highest power of $2$ dividing $n$

L567 wrote: rafaello wrote: Could anyone please tell me briefly what is denoted by $\mu_2(\alpha)$, $\mu_2(\beta)$? Never seen such notation before in this context. It's just $v_2(n)$, which is the highest power of $2$ dividing $n$ aa thanks a lot, xd

L567 wrote: DapperPeppermint wrote: anantmudgal09 wrote: Find all pairs of integers $(a,b)$ so that each of the two cubic polynomials $$x^3+ax+b \, \, \text{and} \, \, x^3+bx+a$$has all the roots to be integers. Proposed by Prithwijit De and Sutanay Bhattacharya Subtract to get $(a-b) x=a-b$. If a=b using vieta find (a+1) (b+1)(c+1)=1, at least one of these is 0 due to sign of product being positive so a =b=0. If a≠b and x=1(both can't hold true), then every root of both the equations must satisfy (a-b) x=a-b which can't be possible if root is not 1 and a-b is not 0 so all roots are 1 which gives contradiction. Wait that doesn't work... U can't just assume they have a common root right? Yea you must be sure about the existence of a common integral root before starting out with this step and I feel that any such claim wouldn't be that useful for solving this problem

ftheftics wrote: Let , $$ P_1(x)=x^3+ax+b \, \, \text{and} \, \,P_2(x)= x^3+bx+a$$ Let, $ b_1, b_2,b_3$ are roots of $P_1$ and $a_1,a_2,a_3$ are roots of $P_2$. so, \[b_1+b_2+b_3 =0 , b_1b_2+b_2b_3+b_3b_1 =a \implies b_1^2 +b_2^2 +b-3^2 =-2a\] clearly atleast one of $b_i$ is divisable by 2 . so , $2\mid b_1b_2b_3 = -b $ , $2 | b$ . similarly , $ 2 | a $ now ,\[2 \mid 0= b_i^3 +ab_i +b \implies 2\mid b_i^3 \implies 8 \mid b_i \] so, $ 8^3 \mid b$ and for $a $ same . using same step get $8^6 \mid a,b$ using infinite descent get , only possibility is $a,b=0$ whoever thinks about , $a,b = 0,-n^6$ forgot about complex numbers . How Can you say$2 |a^3 gives 8|a$ I think your solution is wrong

$\color{red}\textbf{Claim:-}$ The only solution works that is $(a,b)=(0,0).$ $\color{blue}\textbf{Proof:-}$ Simply $x^3=0\implies (a,b)=(0,0)$ Now let's assume the roots of the polynomial $x^3+ax+b$ are $\alpha_1,\beta_1,\gamma_1$ and the roots of the polynomial $x^3+bx+a$ are $\alpha_2,\beta_2,\gamma_2.$ Now By Vieta's Relation form the first polynomial we get, $$\alpha_1+\beta_1+\gamma_1=0$$$$\alpha_1\beta_1+\beta_1\gamma_1+\gamma_1\alpha_1=a$$$$\alpha_1\beta_1\gamma_1=-b$$From first two equations by applying Vieta's Relation we get, $$\alpha_1^2+\beta_1^2+\gamma_1^2=-2a\implies\text{a is negative integer.}$$Now by Vieta's Relation from the second polynomial we get, $$\alpha_2+\beta_2+\gamma_2=0$$$$\alpha_2\beta_2+\beta_2\gamma_2+\gamma_2\alpha_2=b$$$$\alpha_2\beta_2\gamma_2=-a$$From first two equations by applying Vieta's Relation we get, $$\alpha_2^2+\beta_2^2+\gamma_2^2=-2b\implies \text{b is negative integer.}$$From these we conclude that the product of roots of both the polynomials are $+\text{ve.}\implies \alpha_1\beta_1\gamma_1=+\text{ve.}\implies\alpha_2\beta_2\gamma_2=+\text{ve.}$ it also states that among the three roots of the first polynomial $$\alpha_1,\beta_1,\gamma_1\implies\textbf{two are -ve and one is +ve.}$$Now from these there are two cases arise $$\textbf{Case-1:}=\text{One is even and two are odd.}$$$$\textbf{Case-2:}=\text{Three are even.}$$Now Explore $\textbf{Case-1:}\implies\text{One is even and two are odd.}$ we get, $$\sum_{cyc}{}\alpha_1^2=4k_1+1+4k_2+1+4k_3=-2a$$$$4(k_1+k_2+k_3)+2=-2a\implies \text{a comes out to be odd }\implies\textbf{contradiction.}$$Now Explore $\textbf{Case-2:}=\text{Three are even.}$ we get, $$\sum_{cyc}{}\alpha_1^2=-2a\implies 4k_1+4k_2+4k_3=-2a$$$$2(k_1+k_2+k_3)=-a\implies\text{a is even.}$$Now observing $$\alpha_2\beta_2\gamma_2=-a\implies\text{they all have common factor as 2}\implies 8|a, 8|b$$Therefore all the six roots are even and hence $a$ and $b$ are also even. Similarly we get, $$8|\alpha_1,\beta_1,\gamma_1\implies 8|\alpha_2,\beta_2,\gamma_2$$Which leads to an infinite descent, which will not give any further solution. And the claim follows.

Let $P(t)=t^3+at+b$ and $Q(t)=t^3+bt+a$. The discriminant of $P$ is $-4a^3-27b^2$. As all roots of $P$ are real, we must have $$-4a^3-27b^2\geq 0 \qquad (1)$$It follows that $a\leq 0$. Similarly from the discriminant of $Q$ we get $b\leq 0$. If $a=0$ (resp $b=0$) then from $(1)$ we get $b=0$ (resp $a=0$). The pair $(0,0)$ indeed works. Now let us assume $a,b<0$. Let $p=-a>0$, $q=-b>0$. Let nonzero integers $a_1,a_2,a_3$ be the roots of $P(t)=t^3-pt-q$. Then from Vieta's relations \begin{align*} & a_1+a_2+a_3=0 \\ & a_1a_2+a_2a_3+a_3a_1=-p \\ & a_1a_2a_3=b \end{align*}From these equations it is clear that two of the roots must be negative, let us call them $-x$, $-y$. Then the other root is $x+y$. Notice that $$q\left (\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}\right )=-p \implies q\left (\frac{1}{x}+\frac{1}{y}-\frac{1}{x+y}\right )=p \qquad (2)$$Similarly there are integers $u,v>0$ ($-u$, $-v$ are roots of $Q$) so that $$p\left (\frac{1}{u}+\frac{1}{v}-\frac{1}{u+v}\right )=q \qquad (3)$$If $x,y,u,v>1$ then $$\frac{1}{x}+\frac{1}{y}-\frac{1}{x+y}\leq \frac 12+\frac 12-\frac{1}{x+y}=1-\frac{1}{x+y}<1$$and likewise $\frac{1}{u}+\frac{1}{v}-\frac{1}{u+v}<1$. But then $(2)$ implies $p<q$ while $(3)$ yields $p>q$, a contradiction. Now suppose wlog that $x=1$. Then $q=y^2+y$ and $p=q+1=y^2+y+1$ (since $P(-1)=0$). Now, \begin{align*} Q(t)=0 & \implies t^3-(y^2+y)t-(y^2+y+1)=0 \\ & \implies (t+1)(t^2-t+1-y^2-y)=2. \end{align*}From this it is easy to get $y(y+1)=0$ but this is not possible since $y>0$. It follows that the answer is $(0,0)$ only.

This is my favourite problem of INMO '21