exercise in commutative algebra exercise: let $A$ be an integral domain in which all prime ideals are maximal (but may not be Dedekind in general).  let $P_1,  P_2$ be 2 distinct prime ideals.  prove: for any positive integer $m, n$,  $P_1^m+P_2^n=A$. 

so the hint is to prove this first for arbitrary $m$ when $n=1$.  I finished this step by observing that if $P_1^m$ is contained in $P_2$,  then since they re prime ideals $P_1 \subset P_2$.  this contradicts the fact that they re distinguish and maximal.  so this was done. 

but next how to use induction for $m$? I need some hint, thanks!
consider it to be true for n then multiply by P_2 
the RHS is going to be P_2.

see that (P_1)^m*(P_2) + (P_2)^(n+1) is contained in (P_2).

(P_1)^m*(P_2) is contained in (P_1)^m

(P_1)^m is not contained in (P_2)

now let "a" be any such element which is in (P_1)^m but not in (P_2)
note that (a) + (P_2) is contained in (a) + (P_1)^m*(P_2) + (P_2)^(n+1) hence (P_1)^m + (P_2)^(n+1)

and from maximality of (P_2) you get (a) + (P_2) is actually A
consider it to be true for n then multiply by P_2 
the RHS is going to be P_2.

see that (P_1)^m*(P_2) + (P_2)^(n+1) is contained in (P_2).

(P_1)^m*(P_2) is contained in (P_1)^m

(P_1)^m is not contained in (P_2)

now let "a" be any such element which is in (P_1)^m but not in (P_2)
note that (a) + (P_2) is contained in (a) + (P_1)^m*(P_2) + (P_2)^(n+1) hence (P_1)^m + (P_2)^(n+1)

and from maximality of (P_2) you get (a) + (P_2) is actually A