## 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