2001 JBMO Problems/Problem 1

Problem

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

Solution1

Note that for all positive integers $n,$ the value $n^3$ is congruent to $-1,0,1$ modulo $9.$

Proof: set $n = 3k + a.$ Then, $n^3 \equiv 27k^3 + 27k^2a + 9ka^2 + a^3 \equiv a^3 \pmod{9}.$ If $a = 0, n^3 \equiv 0^3 \pmod{9}.$ If $a = 1, n^3 \equiv 1^3 \pmod{9}.$ If $a = 2, n^3 \equiv 2^3 \equiv -1 \pmod{9}.$

Since $2001 \equiv 3 \pmod{9},$ we find that $a^3,b^3,c^3 \equiv 1 \pmod{9}.$ Thus, $a,b,c \equiv 1 \pmod{3},$ and the only numbers congruent to $1$ modulo $3$ are $1,4,7,10.$

WLOG, let $a \ge b \ge c.$ That means $a^3 \ge b^3, c^3$ and $3a^3 \ge 2001.$ Thus, $a^3 \ge 667,$ so $a = 10.$

Now $b^3 + c^3 = 1001.$ Since $b^3 \ge c^3,$ we find that $2b^3 \ge 1001.$ That means $b = 10$ and $c = 1.$

In summary, the only solutions are $\boxed{(10,10,1),(10,1,10),(1,10,10)}.$

Solution2

You can also watch remainders modulo 7 which are also -1,0,1. The rest is almost identical as in solution 1.

2001 JBMO (Problems • Resources)
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2001 JBMO Problems/Problem 1

Contents

Problem

Solution1

Solution2

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