pigor: ...., zauważ , że jeśli m(n)=n
3−3n
2+n+2 i n∊N
+, to
m(2)=0 i m(n)= n
3−2n
2−n
2+2n−n+2= n
2(n−2)−n(n−2)−1(n−2)= (n−2)(n
2−n−1),
zatem U{2(n−2){n
3−3n
2+n+2}= U{2(n−2){(n−2)(n
2−n−1)}∊ C i n−2≠0 ⇔
⇔ (n
2−n−1= −1 ∨ n
2−n−1= 1 ∨ n
2−n−1= −2 ∨ n
2−n−1= 2) ∧ (*)
n∊N+\{2} ⇒
⇒ n
2−n= 0 ∨ n
2−n−2= 0 ∨ n
2−n+1= 0 ∨ n
2−n−3= 0 ⇔
⇔ n(n−1)= 0 ∨ (n−2)(n+1)= 0 ∨ n ∊∅ ∨ n ∉ N stąd i z (*)
n∊{1} . ...