 

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March 2010 Difference of two square roots


Is every positive integer power of
consecutive integers?

2  1
2
2 1  9  8

3
 50  49

2  1 the difference between the square roots of
Solution
This can be proved by induction, taking even powers and odd powers separately.
Even powers.


2
2 1  9  8

Assume
Then


2 1

2 1
2k
2k 2
[ or the starting point could be


0
2 1  1  0 ]
 A  B 2  A2  2 B 2 where A2  2 B 2  1
 2  1  2  1
  3  2 2  A  B 2 

2
2k
  3 A  4 B   2  2 A  3B 

3 A  4B 
2
 2  2 A  3B 
2
Now  3 A  4 B   2  2 A  3B   A2  2 B 2  1
2
So


2 1
2
2k
takes the desired form.
Odd powers.


1
2 1  2  1
Assume
Then



2 1

2 1
2 k 1
2 k 1
 C 2  D  2C 2  D 2 where 2C 2  D 2  1
 2  1  2  1
  3  2 2  C 2  D 

2 k 1
2
 2  3C  2 D    4C  3D 
Now 2  3C  2 D    4C  3D   2C 2  D 2  1
2
So


2 1
2
2 k 1
also takes the desired form.
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