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Difference between revisions of "2001 JBMO Problems/Problem 1"

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Solve the equation <math>a^3 + b^3 + c^3 = 2001</math> in positive integers.
 
Solve the equation <math>a^3 + b^3 + c^3 = 2001</math> in positive integers.
  
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==Solution==
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==Solution1==
  
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Note that for all positive integers <math>n,</math> the value <math>n^3</math> is congruent to <math>-1,0,1</math> [[modulo]] <math>9.</math> Since <math>2001 \equiv 3 \pmod{9},</math> we find that <math>a^3,b^3,c^3 \equiv 1 \pmod{9}.</math> Thus, <math>a,b,c \equiv 1 \pmod{3},</math> and the only numbers congruent to <math>1</math> modulo <math>3</math> are <math>1,4,7,10.</math>
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Note that for all positive integers <math>n,</math> the value <math>n^3</math> is congruent to <math>-1,0,1</math> [[modulo]] <math>9.</math>
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Proof: set <math>n = 3k + a.</math> Then, <math>n^3 \equiv 27k^3 + 27k^2a + 9ka^2 + a^3 \equiv a^3 \pmod{9}.</math> If <math>a = 0, n^3 \equiv 0^3 \pmod{9}.</math> If <math>a = 1, n^3 \equiv 1^3 \pmod{9}.</math> If <math>a = 2, n^3 \equiv 2^3 \equiv -1 \pmod{9}.</math>
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Since <math>2001 \equiv 3 \pmod{9},</math> we find that <math>a^3,b^3,c^3 \equiv 1 \pmod{9}.</math> Thus, <math>a,b,c \equiv 1 \pmod{3},</math> and the only numbers congruent to <math>1</math> modulo <math>3</math> are <math>1,4,7,10.</math>
  
 
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Now <math>b^3 + c^3 = 1001.</math>  Since <math>b^3 \ge c^3,</math> we find that <math>2b^3 \ge 1001.</math>  Thus, <math>b = 10</math> and <math>c = 1.</math>
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Now <math>b^3 + c^3 = 1001.</math>  Since <math>b^3 \ge c^3,</math> we find that <math>2b^3 \ge 1001.</math>  That means <math>b = 10</math> and <math>c = 1.</math>
  
 
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In summary, the only solutions are <math>\boxed{(10,10,1),(10,1,10),(1,10,10)}.</math>
 
In summary, the only solutions are <math>\boxed{(10,10,1),(10,1,10),(1,10,10)}.</math>
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==Solution2==
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You can also watch remainders modulo 7 which are also -1,0,1. The rest is almost identical as in solution 1.
  
 
==See Also==
 
==See Also==

Latest revision as of 09:53, 30 April 2025

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.

See Also

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