Răspuns:
a-1,a,a+1 - nr consecutive
(a-1)²+a²+(a+1)²=a²-2a+1+a²+a²+2a+1=3a²+2 - suma patratelor
D = I ·C + R -teorema impartirii cu rest
(3a²+2):3=3a²:3+2:3=a²+2:3
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luam trei nr consecutive 3,4,5
3²=9
4²=16
5²=25
9+16+25=50
50:3=16 rest 2