No.
Let m=(N1); then N^2+N1=(m+1)^2+m, but every prime factor of (m+1)^2+m is congruent to {0, 1, 4} mod 5 (Sloane A038872) and so must end in {1, 5, 9}.
As to compound divisors:
1*1=1, 1*5=5, 1*9=9, 5*5=25, 5*9=45, 9*9=81; any combination of terminal digits produces a new digit in the same set {1,5,9}, so 3 and 7 can never be terminal digits of any divisor of N^{2} + N  1.
Edited on January 11, 2015, 2:37 am

Posted by broll
on 20150111 02:35:48 