2001 JBMO Problems/Problem 1

Problem

Solve the equation $a^3 + b^3 + c^3 = 2001$ in positive integers.

Solution

Note that for all positive integers $n,$ the value $n^3$ is congruent to $-1,0,1$ modulo $9.$ Since $2001 \equiv 3 \pmod{9},$ we find that $a,b,c \equiv 1 \pmod{9}.$

The only numbers congruent to $1$ modulo $9$ that are also less than $2001$ are $10$ and $1.$ With some quick checking, we find that the only solutions are $\boxed{(10,10,1),(10,1,10),(1,10,10)}.$

2001 JBMO (Problems • Resources)
Preceded by First Problem	Followed by Problem 2
1 • 2 • 3 • 4
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2001 JBMO Problems/Problem 1

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