Numerele consecutive pare notam cu:
2k ; 2k+2 ; 2k+4 si 2k+6
⇵
(2k+6)³ - 2k*(2k+2)(2k+4) obtinem un numar divizibil cu 8 !!!
8k³+72k²+216k+216 - (4k²+4k)(2k+4)=
= 8k³+72k²+216k+216 - 8k³- 8k²- 16k²-16k=
= 48k²-200k+216 = 8*(6k² - 25k + 27) divizibil cu 8 !!!!