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Power Expressions and Division (Posted on 2016-02-06) Difficulty: 3 of 5
Determine all possible pairs (X, M) of positive integers such that:

XM+2M+1 divides XM+1+2M+1+1

No Solution Yet Submitted by K Sengupta    
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Just one? Comment 1 of 1
For easyness:
D(m+1)=1+2^(m+1)+x^(m+1)
D(m)=1+2^m+x^m

See that for x>1:
D(m+1)<x*D(m)

As: x^(m+1)=x*x^m=(x-1)*x^m+x^m, and same for 2^m+1
it's possible to express D(m+1) as a function of D(m)

D(m+1)=D(m)+2^m+(x-1)x^m

doing it again:
D(m+1)=2*D(m)+(x-2)*(x^m)-1

Doing it (x-1) times
D(m+1)=(x-1)*D(m)+x(x^(m-1)-2^m-1)+3*2^m+2

Doing it x times is unnecessary as D(m+1)<x*D(m)

The expression on the right of D(m) adds (x^m + 3*2^m+2) and detract (2^m*x+x). This is going to be always positive for the presence of x^m in the addition, except if m=1. [This implies that usually (x-1)*D(m)<D(m+1)<x*D(m)].
But if m=1, for x{4,5}, the second member is igual or higher and the expression is negative. Concretely for x=4, m=1 the expression is 0, and so verify that D(m+1)=(x-1)*D(m) 

D(m) 1+2+4=7
D(m+1)=1+4+16=21

Edited on February 9, 2016, 4:15 am
  Posted by armando on 2016-02-08 17:10:31

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