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I would appreciate if somebody could help me with the following problem.

What is the number of solutions to the equation $$2024 = a^3+b^3+c^3-3abc$$ with integers $a\geq b\geq c$?

My work: I tried to approach it through factorization, but it didn't work. \begin{align*} 2024 &=a^3+b^3+c^3-3abc\\ &=(a+b+c)\left\{a^2+b^2+c^2 -ab-bc-ca \right\}\\ &=(a+b+c)\left\{(a+b+c)^2-3(ab+bc+ca) \right\}\\ &=(a+b+c)\left\{ \frac{1}{2}(a-b)^2+ \frac{1}{2}(b-c)^2+\frac{1}{2}(c-a)^2 \right\} \end{align*} The existence of roots was found using mathematica11. {a, b, c} /. Solve[a^3 + b^3 + c^3 - 3 a b c == 2024 && a <= b <= c, {a, b, c}, Integers] {{168, 168, 170}, {674, 675, 675}}

Young
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1 Answers1

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The factorization is a very good start.

The question doesn't say if any of them can be negative, but from the last expression, we can confirm that $a+b+c>0$. So when we factorize 2024, we only need to consider the positive factors.

Noticing that in the third expression, we can tell that the second factor modulo 3 is $(a+b+c)^2$, which must be 0 or 1 $\mod 3$. Since $2024\equiv 2 \mod 3$, we must have $a+b+c\equiv 2 \mod 3$. The only possible ways to factor 2024 are:

a+b+c=  a^2+b^2+c^2-ab-bc-ca=
2       1012
8       253
11      184
23      88
44      46
92      22
506     4
2024    1

Next, let's write $a-b=p, b-c=q$. We still don't know if $b$ and $c$ can be negative, but we are sure that $p$ and $q$ are non negative since $a\geq b \geq c$. So the second factor becomes $p^2+q^2+pq$. Let's find the solutions for the factor on the right.

Noticing that if $p^2+q^2+pq$ is even, then both $p$ and $q$ must be even, in which case it must be a multiple of 4. That means we can divide $p,q$ by 2 and the result by 4, and it must still be a solution. Which implies that the factorization of the result must have an even power of 2. That rules out the middle 4 cases, and we're only left with

2    1012
8    253
506  4
2024 1

The solution of $p^2+q^2+pq=1$ is $p=0,q=1$ or $p=1,q=0$. Noticing that $a+b+c$ becomes $p+2q+3c$, we must have $p+2q\equiv 2\mod 3$, so $p=0,q=1$. And we found the solution {675, 675, 674}. Similar for $p^2+q^2+pq=4$ which we can divide $p$ and $q$ by 2 and it reduces to the previous problem. The solution is $p=2,q=0$. And we found the solution {170, 168, 168}.

To solve $p^2+q^2+pq=253$, I can't think of any quick shortcuts, so we have to do it the hard way. we need $q^2-4(q^2-253)$ to be a square, and $q\leq 15$. Due to the symmetry of the problem, we only need to check when $p\leq q$, so we only need to check $q\geq 10$. After we rule out all the cases $10\leq q\leq 15$, we are sure that there's no solution in this case. This is the only tedious part of the proof.

For the case $p^2+q^2+pq=1012$, we just divide $p$ and $q$ by two, and it reduces to the previous problem which we know have no solutions. So the only two solutions are {170, 168, 168} and {675, 675, 674}.

Yi Jiang
  • 464