Partitioning integer N into 3 positive integers

 
 
Let   P_3 \, (n)   denote the number of ways a positive integer   n   can be partitioned into 3 positive integers.

Prove that if   a, \; b, \; c   are positive integers and   a^2 \; + \; b^2 \; = \; c^2,

then   P_3 \, (a) \; + \; P_3 \, (b) \; = \; P_3 \, (c)

 
 
Let’s take, for example, the Primitive Pythagorean triples
(3, \; 4, \; 5)        (5, \; 12, \; 13)

3^2 \; + \; 4^2 \; = \; 5^2                 5^2 \; + \; 12^2 \; = \; 13^2

Also,

P_3 \, (3) \; = \; 1                 P_3 \, (4) \; = \; 12                 P_3 \, (5) \; = \; 2

P_3 \, (12) \; = \; 12              P_3 \, (13) \; = \; 14

giving us

P_3 \, (3) \; + \; P_3 \, (4) \; = \; P_3 \, (5)

P_3 \, (5) \; + \; P_3 \, (12) \; = \; P_3 \, (13)
 
 

P_3 \, (3) \; = \; 1
1 + 1 + 1 = 3

P_3 \, (4) \; = \; 1
2 + 1 + 1 = 4

P_3 \, (5) \; = \; 2
3 + 1 + 1 = 5 ………. 2 + 2 + 1 = 5

P_3 \, (12) \; = \; 12
10 + 1 + 1 = 12 ………. 9 + 2 + 1 = 12 ………. 8 + 3 + 1 = 12
  8 + 2 + 2 = 12 ………. 7 + 4 + 1 = 12 ………. 7 + 3 + 2 = 12
  6 + 5 + 1 = 12 ………. 6 + 4 + 2 = 12 ………. 6 + 3 + 3 = 12
  5 + 5 + 2 = 12 ………. 5 + 4 + 3 = 12 ………. 4 + 4 + 4 = 12

P_3 \, (13) \; = \; 14
11 + 1 + 1 = 13 ………. 10 + 2 + 1 = 13 ………. 9 + 3 + 1 = 13
  9 + 2 + 2 = 13 ………. 8 + 4 + 1 = 13 ………. 8 + 3 + 2 = 13
  7 + 5 + 1 = 13 ………. 7 + 4 + 2 = 13 ………. 7 + 3 + 3 = 13
  6 + 6 + 1 = 13 ………. 6 + 5 + 2 = 13 ………. 6 + 4 + 3 = 13
  5 + 5 + 3 = 13 ………. 5 + 4 + 4 = 13

 
 
 

 
 
 
 
 
 
 
 
 
 
 
 
 
 

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About benvitalis

math grad - Interest: Number theory
This entry was posted in Number Puzzles and tagged . Bookmark the permalink.

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