For k=1,2:
+) If m≡1(mod 4) then choose n≡1(mod 4)
+) If m≡3(mod 4) then choose n≡3(mod 4)
We define n(k,m):n(k,m)n(k,m)−m⋮2k
From above we solve for k=1,2
Now we will induction, for any m, assume exist n(k,m). We will prove exist n(k+1,m)
+) If n(k,m)n(k,m)−m⋮2k+1 then choose n(k+1,m)=n(k,m)
+) If n(k,m)n(k,m)−m≡2k(mod 2k+1). We will choose n(k+1,m) in the form a:n(k+1,m)=a.2k+n(k,m)
We need (a.2k+n(k,m))a.2k+n(k,m)−m⋮2k+1, but (a.2k+n(k,m))a.2k−1⋮2k+1 (LTE)
So we need (a.2k+n(k,m))n(k,m)−m⋮2k+1
Apply Newton"s binomial, we neew to choose a:n(k,m).n(k,m)n(k,m)−1.a2k+n(k,m)n(k,m)−m⋮2k+1
Which mean 2k.(n(k,m).n(k,m)n(k,m)−1.a+1)⋮2k+1. Just choose a odd now
Q.E.D