Notam astfel
a;b∈N unde
a>b si a=b+1
aaa+333²=100a+10a+a+333²
100a+10a+a=100(b+1)+10(b+1)+b+1=(b+1)(100+10+1)=(b+1)·111
111bbb=100000+10000+1000+100b+10b+b=111000+100b+10b+b
111000+100b+10b+b-111(b+1)=
111000+111b-111b-111=
111000-111=
110889=333·333=333²⇒adevarat⇒daca a si b sunt cifre consecutive in baza 10 cu a>b atunci aaa+333²=111bbb.
