Solution
We find out what are the integer cubes congruent to modulo \(9\):
- if \(n \equiv 0\pmod 9\), then \(n^{3} \equiv 0^{3} \equiv 0\pmod 9\);
- If \(n \equiv 1\pmod 9\), \(n^{3} \equiv 1^{3} \equiv 1\pmod 9\);
- 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 \(n^{3} \equiv(-4)^{3} \equiv-1\pmod 9\);
- 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 to \(\pm 1\) modulo \(9\).
Let us find out what can the sum of
two integer cubes be congruent to modulo \(9\).
| \(a^{3}\pmod 9\) |
-1 |
0 |
1 |
| \(b^{3}\pmod 9\) |
|
|
|
| -1 |
-2 |
-1 |
0 |
| 0 |
-1 |
0 |
1 |
| 1 |
0 |
1 |
2 |
We conclude that the sum of two integer cubes modulo \(9\)
can equal to any value from \(-2,-1,0,1,2\),
but nothing else. Since \(2022 \equiv 6 \equiv-3 \pmod 9\) does not equal any of these values,
the sum of two integer cubes cannot equal \(2022\).
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 \(n^{3} \equiv(-4)^{3} \equiv-1\pmod 9\);
- 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 skaitllu kubu summa pēc moduļa \(9\).
| \(a^{3}\pmod 9\) |
-1 |
0 |
1 |
| \(b^{3}\pmod 9\) |
|
|
|
| -1 |
-2 |
-1 |
0 |
| 0 |
-1 |
0 |
1 |
| 1 |
0 |
1 |
2 |
Esam ieguvuši, ka divu šādu skaitlu summa pēc moduļa \(9\) var
pieņemt jebkuru no vērtībām \(-2,-1,0,1,2\), taču nekādas citas.
Tā kā \(2022 \equiv 6 \equiv-3 \pmod 9\) neparādās starp šīm vērtībām,
tad divu veselu skaitļu kubu summa nevar būt \(2022\).