Solution
We first find out what the integer cubes are congruent to modulo \(9\):
- if \(n \equiv 0 \pmod 9\), then \(n^{3} \equiv 0^{3} \equiv 0 \pmod 9\);
- if \(n \equiv 1 \pmod 9\), then \(n^{3} \equiv 1^{3} \equiv 1 \pmod 9\);
- if \(n \equiv 2 \pmod 9\), then \(n^{3} \equiv 2^{3} \equiv 8 \equiv-1 \pmod 9\);
- if \(n \equiv 3 \pmod 9\), then \(n^{3} \equiv 3^{3} \equiv 27 \equiv 0 \pmod 9\);
- if \(n \equiv 4 \pmod 9\), then \(n^{3} \equiv 4^{3} \equiv 64 \equiv 1 \pmod 9\);
- if \(n \equiv 5 \equiv-4 \pmod 9\), then \(n^{3} \equiv(-4)^{3} \equiv-4^{3} \equiv-1 \pmod 9\);
- if \(n \equiv 6 \equiv-3 \pmod 9\), then \(n^{3} \equiv(-3)^{3} \equiv 0 \pmod 9\);
- if \(n \equiv 7 \equiv-2 \pmod 9\), then \(n^{3} \equiv(-2)^{3} \equiv 1 \pmod 9\);
- if \(n \equiv 8 \equiv-1 \pmod 9\), then \(n^{3} \equiv(-1)^{3} \equiv-1 \pmod 9\).
So integer cubes are congruent to \(0\) or \(\pm 1\) modulo \(9\).
Let us consider what is the sum of two integer cubes congruent to modulo \(9\):
| \(a^{3} \pmod 9\) |
\(-1\) |
\(0\) |
\(1\) |
| \(b^{3} \pmod 9\) |
\(-2\) |
\(-1\) |
\(0\) |
| \(-1\) |
\(-1\) |
\(0\) |
\(1\) |
| \(0\) |
\(0\) |
\(1\) |
\(2\) |
| \(1\) |
|
|
|
Next, consider what is the sum of three integer cubes modulo \(9\).
| \(a^{3}+b^{3} \pmod 9\) |
\(-1\) |
\(0\) |
\(1\) |
\(-2\) |
\(2\) |
| \(-1\) |
\(-2\) |
\(-1\) |
\(0\) |
\(-3\) |
\(1\) |
| \(0\) |
\(-1\) |
\(0\) |
\(1\) |
\(-2\) |
\(2\) |
| \(1\) |
\(0\) |
\(1\) |
\(2\) |
\(-1\) |
\(3\) |
We conclude that the sum of three integer cubes modulo \(9\) can take any of the values
\(-3 ;-2 ;-1 ; 0 ; 1 ; 2 ; 3\) and nothing else.
As \(2023^{2} \equiv 7^{2} \equiv 4 \equiv-5 \pmod 9\) does not appear among these values,
the sum of three whole skaitļu cubes cannot be \(2023^{2}\).
Atrisinājums
Vispirms noskaidrosim, ar ko var būt kongruenti veselu skaitlu kubi pēc moduļa \(9\):
- ja \(n \equiv 0 \pmod 9\), tad \(n^{3} \equiv 0^{3} \equiv 0 \pmod 9\);
- ja \(n \equiv 1 \pmod 9\), tad \(n^{3} \equiv 1^{3} \equiv 1 \pmod 9\);
- ja \(n \equiv 2 \pmod 9\), tad \(n^{3} \equiv 2^{3} \equiv 8 \equiv-1 \pmod 9\);
- ja \(n \equiv 3 \pmod 9\), tad \(n^{3} \equiv 3^{3} \equiv 27 \equiv 0 \pmod 9\);
- ja \(n \equiv 4 \pmod 9\), tad \(n^{3} \equiv 4^{3} \equiv 64 \equiv 1 \pmod 9\);
- ja \(n \equiv 5 \equiv-4 \pmod 9\), tad \(n^{3} \equiv(-4)^{3} \equiv-4^{3} \equiv-1 \pmod 9\);
- ja \(n \equiv 6 \equiv-3 \pmod 9\), tad \(n^{3} \equiv(-3)^{3} \equiv 0 \pmod 9\);
- ja \(n \equiv 7 \equiv-2 \pmod 9\), tad \(n^{3} \equiv(-2)^{3} \equiv 1 \pmod 9\);
- ja \(n \equiv 8 \equiv-1 \pmod 9\), tad \(n^{3} \equiv(-1)^{3} \equiv-1 \pmod 9\).
Tātad veselu skaitlu kubi ir kongruenti ar 0 vai \(\pm 1\) pēc moduļa 9. Aplūkosim, ar ko var būt kongruenta divu veselu skaitlu kubu summa pēc moduļa 9.
| \(a^{3} \pmod 9\) |
-1 |
0 |
1 |
| \(b^{3} \pmod 9\) |
-2 |
-1 |
0 |
| -1 |
-1 |
0 |
1 |
| 0 |
0 |
1 |
2 |
| 1 |
|
|
|
Tagad aplūkojam, ar ko var būt kongruenta trīs veselu skaitļu kubu summa pēc modula 9.
| \(a^{3}+b^{3} \pmod 9\) |
-1 |
0 |
1 |
-2 |
2 |
| -1 |
-2 |
-1 |
0 |
-3 |
1 |
| 0 |
-1 |
0 |
1 |
-2 |
2 |
| 1 |
0 |
1 |
2 |
-1 |
3 |
Esam ieguvuši, ka trīs šādu skaițu summa pēc moduļa 9 var pienemt jebkuru no vērtībām \(-3 ;-2 ;-1 ; 0 ; 1 ; 2 ; 3\) un nekādas citas. Tā kā \(2023^{2} \equiv 7^{2} \equiv 4 \equiv-5 \pmod 9\) neparādās starp šīm vērtībām, tad trīs veselu skaitļu kubu summa nevar būt \(2023^{2}\).