Explicație pas cu pas:
5/a€N <=> a|5 => a€D5 in N =>
a€{1,5}
3a+1/a€N <=> a|3a+1
a|3a+1
a|a => a|3a
=> a|3a+1-3a => a|1 => a€D1 in N => a€{1}
2a+7/a+1€N <=> a+1|2a+7
a+1|2a+7
a+1|a+1=>a+1|2a+2
=> a+1|2a+7-2a-2 => a+1|5 => a+1€D5 in N => a+1€{1,5} =>
a€{0,4}
