There is a simple formula that gives all the Pythagorean triples.

Suppose that *m* and *n* are two positive integers,
with *m* < *n*.
Then *n*^{2} - *m*^{2}, 2*mn*, and
*n*^{2} + *m*^{2} is a Pythagorean triple.

It's easy to check algebraically that the sum of the squares of the first
two is the same as the square of the last one. Is it the case that
*every*
triple can be generated in this manner?

Here are the first few triples for *m* and *n* between 1 and 10.
Notice any patterns?

| m= 1 2 3 4 5 6 7 8 9 ---+------------------------------------------------------------------------------------------------------------------ n= | 2 | [3,4,5] | 3 | [8,6,10] [5,12,13] | 4 | [15,8,17] [12,16,20] [7,24,25] | 5 | [24,10,26] [21,20,29] [16,30,34] [9,40,41] | 6 | [35,12,37] [32,24,40] [27,36,45] [20,48,52] [11,60,61] | 7 | [48,14,50] [45,28,53] [40,42,58] [33,56,65] [24,70,74] [13,84,85] | 8 | [63,16,65] [60,32,68] [55,48,73] [48,64,80] [39,80,89] [28,96,100] [15,112,113] | 9 | [80,18,82] [77,36,85] [72,54,90] [65,72,97] [56,90,106] [45,108,117] [32,126,130] [17,144,145] | 10 | [99,20,101] [96,40,104] [91,60,109] [84,80,116] [75,100,125] [64,120,136] [51,140,149] [36,160,164] [19,180,181]