this is a brilliant ques indeed.
we have
a+d= mp
(b-1)c= mq
and ab-a+c= mr
now abn+cn+d= abn+cn+mp – a
= a(bn – 1)+cn+mp
now, put b – 1= k
so that kc= mq and ak+c= mr
also, a(bn – 1)+cn+mp= a((k+1)n – 1)+cn+mp
= a{(1+nk+nC2*k^2+....+nCn*k^n) – 1}+cn+mp [using binomial expansion]
= n(ak+c)+ak2(nC2+....+nCn*kn-2)+mp [rearranging the terms]
=nmr+mp+ak2(nC2+....+nCn*kn-2)
now consider ak+c= mr
multiply both sides by k
ak^2 + ck= mrk
but ck= mq
so ak^2 = mrk – mq= m(rk – q)
hence, abn+cn+d = nmr+mp+ak2(nC2+....+nCn*kn-2)= =nmr+mp+m(rk-q)(nC2+....+nCn*kn-2)
= m*(nr+p+(rk-q)(nC2+....+nCn*kn-2)), which is divisible by m
