IMO 2019 Shortlisted Problems

Solution 1. We will start by proving that $c=1$. Note that $$3a^3⩾a^3+b^3+c^3>a^3$$ So $3a^3⩾(abc{)}^2>a^3$ and hence $3a⩾b^2{c}^2>a$. Now $b^3+c^3=a^2(b^2{c}^2-a)⩾a^2$, and so $$18b^3⩾9(b^3+c^3)⩾9a^2⩾b^4{c}^4⩾b^3{c}^5$$ so $18⩾c^5$ which yields $c=1$.

Now, note that we must have $a>b$, as otherwise we would have $2b^3+1=b^4$ which has no positive integer solutions. So $$a^3-b^3⩾(b+1{)}^3-b^3>1$$ and $$2a^3>1+a^3+b^3>a^3,$$ which implies $2a^3>a^2{b}^2>a^3$ and so $2a>b^2>a$. Therefore $$4(1+b^3)=4a^2(b^2-a)⩾4a^2>b^4$$ so $4>b^3(b-4)$; that is, $b⩽4$.

Now, for each possible value of $b$ with $2⩽b⩽4$ we obtain a cubic equation for $a$ with constant coefficients. These are as follows: $$\begin{array}{ll}b=2:& a^3-4a^2+9=0\\ b=3:& a^3-9a^2+28=0\\ b=4:& a^3-16a^2+65=0.\end{array}$$ The only case with an integer solution for $a$ with $b⩽a$ is $b=2$, leading to $(a,b,c)=(3,2,1)$.

Solution 2. Again, we will start by proving that $c=1$. Suppose otherwise that $c⩾2$. We have $a^3+b^3+c^3⩽3a^3$, so $b^2{c}^2⩽3a$. Since $c⩾2$, this tells us that $b⩽\sqrt{3a}/4$. As the right-hand side of the original equation is a multiple of $a^2$, we have $a^2⩽2b^3⩽2(3a/4{)}^{3/2}$. In other words, $a⩽\frac{27}{16}<2$, which contradicts the assertion that $a⩾c⩾2$. So there are no solutions in this case, and so we must have $c=1$.

Now, the original equation becomes $a^3+b^3+1=a^2{b}^2$. Observe that $a⩾2$, since otherwise $a=b=1$ as $a⩾b$. The right-hand side is a multiple of $a^2$, so the left-hand side must be as well. Thus, $b^3+1⩾a^2$. Since $a⩾b$, we also have $$b^2=a+\frac{b^3+1}{a^2}⩽2a+\frac{1}{a^2}$$ and so $b^2⩽2a$ since $b^2$ is an integer. Thus $(2a{)}^{3/2}+1⩾b^3+1⩾a^2$, from which we deduce $a⩽8$.

Now, for each possible value of $a$ with $2⩽a⩽8$ we obtain a cubic equation for $b$ with constant coefficients. These are as follows: $$\begin{array}{ll}a=2:& b^3-4b^2+9=0\\ a=3:& b^3-9b^2+28=0\\ a=4:& b^3-16b^2+65=0\\ a=5:& b^3-25b^2+126=0\\ a=6:& b^3-36b^2+217=0\\ a=7:& b^3-49b^2+344=0\\ a=8:& b^3-64b^2+513=0.\end{array}$$ The only case with an integer solution for $b$ with $a⩾b$ is $a=3$, leading to $(a,b,c)=(3,2,1)$.

Solution 3. Set $k=(b^3+c^3)/a^2⩽2a$, and rewrite the original equation as $a+k=(bc{)}^2$. Since $b^3$ and $c^3$ are positive integers, we have $(bc{)}^3⩾b^3+c^3-1=k{a}^2-1$, so $$a+k⩾(k{a}^2-1{)}^{2/3}$$ As in Comment 1.2, $k$ is a positive integer; for each value of $k⩾1$, this gives us a polynomial inequality satisfied by $a$: $$k^2{a}^4-a^3-5k{a}^2-3k^{2}a-(k^3-1)⩽0.$$ We now prove that $a⩽3$. Indeed, $$0⩾\frac{k^2{a}^4-a^3-5k{a}^2-3k^{2}a-(k^3-1)}{k^2}⩾a^4-a^3-5a^2-3a-k⩾a^4-a^3-5a^2-5a,$$ which fails when $a⩾4$. This leaves ten triples with $3⩾a⩾b⩾c⩾1$, which may be checked manually to give $(a,b,c)=(3,2,1)$.

Solution 4. Again, observe that $b^3+c^3=a^2(b^2{c}^2-a)$, so $b⩽a⩽b^2{c}^2-1$. We consider the function $f(x)=x^2(b^2{c}^2-x)$. It can be seen that that on the interval $[0,b^2{c}^2-1]$ the function $f$ is increasing if $x<\frac{2}{3}{b}^2{c}^2$ and decreasing if $x>\frac{2}{3}{b}^2{c}^2$. Consequently, it must be the case that $$b^3+c^3=f(a)⩾\min \left(f(b),f(b^2{c}^2-1)\right)$$ First, suppose that $b^3+c^3⩾f(b^2{c}^2-1)$. This may be written $b^3+c^3⩾(b^2{c}^2-1{)}^2$, and so $$2b^3⩾b^3+c^3⩾(b^2{c}^2-1{)}^2>b^4{c}^4-2b^2{c}^2⩾b^4{c}^4-2b^3{c}^4$$ Thus, $(b-2)c^4<2$, and the only solutions to this inequality have $(b,c)=(2,2)$ or $b⩽3$ and $c=1$. It is easy to verify that the only case giving a solution for $a⩾b$ is $(a,b,c)=(3,2,1)$.

Otherwise, suppose that $b^3+c^3=f(a)⩾f(b)$. Then, we have $$2b^3⩾b^3+c^3=a^2(b^2{c}^2-a)⩾b^2(b^2{c}^2-b)$$ Consequently $b{c}^2⩽3$, with strict inequality in the case that $b\ne c$. Hence $c=1$ and $b⩽2$. Both of these cases have been considered already, so we are done.