Pythagorean triples
Problem : Find all triples of integers (a,b,c)
such that a^{2}+b^{2}=c^{2}.
Prerequisites :
If a=ua' and b=ub' then c^{2}=u^{2}(a'^{2}+b'^{2}).
As a'^{2}+b'^{2} is an integer, c
is a multiple of u.
Thus we can restrict the study to primitive Pythagorean triples,
that is when a and b are coprimes. It is equivalent
to say that a and c are coprimes.
Assume a is odd. Then a^{2}1=(a+1)(a1)
is multiple of 8.
If b was also odd then c^{2}2 would be a multiple
of 8, which is impossible, as c^{2} must either be
odd or multiple of 4.
Thus b is even, c is odd, and (c+a), (ca)
are even.
Let u=(c+a)/2 , v=(ca)/2,
so that c=u+v and a=uv
u and v are coprimes, because a and c
are coprimes (any common divisor of u and v would be
a common divisor of a and c). Writing b=2b'
the relation a^{2}+b^{2}=c^{2}
gives b'^{2}= (c^{2}a^{2})/4= uv.
uv is a square, thus has only even exponents in its prime
factorization.
u,v are coprimes, thus have no common element in their
prime factorization.
Thus u and v also each have only even exponents in
their factorization. Thus both are squares.
In conclusion, u=m^{2} and v=n^{2},
so that all solutions are of the form
a = m^{2}n^{2}
b = 2mn
c = m^{2}+n^{2}
Conversely, any triple (a,b,c) defined in this
way from m,n, satisfies a^{2}+b^{2}=c^{2}.
As a and c must be odd, we must have either m
even and n odd, or m odd and n even. Both
cases can be exchanged by exchanging m and n, but
this changes the sign of a. So we may reduce the study to
the case when m is even and n is odd, if we don't
forget that some solutions use the case n>m that
appear with a<0.
The above formulas can be written in a compact form using complex
numbers: a+ib = (m+in)^{2}
If, on the other hand, we take m and n to be both
odd, then we have m+in = (1+i)(m'+in')
for some relative integers m' and n' where one is odd and the other
is even. Then,
a+ib = (m+in)^{2}=2i(m'+in')^{2}
= 2ia'  2b' where a'+ib' = (m'+in')^{2}.
Other proof
An even shorter proof exists but makes use of another mathematical
result : that the existence and uniqueness of prime factorization
does not only apply to the set of integers, but also to the Gaussian
integers (which form a principal ideal domain, and thus a unique factorization domain).
So the number (a+ib)(aib)=c^{2}
has a unique factorization into Gaussian prime numbers. When the
triple (a,b,c) is primitive, the number a+ib
is not divisible by any nonunit integer, and thus cannot have any
pair of conjugate prime factors. As the prime factors of aib
are the conjugate of those of a+ib, the numbers a+ib
and aib are coprimes. But their product is a
square, thus each one is also a square, up to a unit
factor.
List of smallest Pythagorean triples
Here is the list of first Pythagorean triples (a,b,c)
given by a+ib = (m+in)^{2}:
n \ m

0

2

4

6

8

10

1

1,0,1

3,4,5

15,8,17

35,12,37

63,16,65

99,20,101

3

9,0,9

5,12,13

7,24,25

27,36,45

55,48,73

91,60,109

5

25,0,25

21,20,29

9,40,41

11,60,61

39,80,89

75,100,125

7

49,0,49

45,28,53

33,56,65

13,84,85

15,112,113

51,140,149

9

81,0,81

77,36,85

65,72,97

45,108,117

17,144,145


We notice that 65 is the hypotenuse of 2 triangles in this list :
33+56i=(4+7i)^{2} , and 63+16i=(8+i)^{2}. This is
because we have the multiple factorization
65=(4+7i)(47i)=(8+i)(8i)=5×13
where
5=(2+i)(2i)
13=(3+2i)(32i)
8+i=(3+2i)(2i)
4+7i=(3+2i)(2+i)
The same for 85 = 5×17 :
(2+i)(4+i)=7+6i
(2+i)(4i)=9+2i
Another parametrization
We may reject the choice of (m even, n odd) as
artificial, and consider that it should be mixed with (m odd,
n even) in the same table. However we need to still present
this as a table with ligns and columns. If we kept m and n
as parameters we would have twice more cells, with the cases of
(even, even) and (odd, odd) giving values of nonprimitive
Pythagorean triples:
n \ m

0

1

2

3

4

5

6

0

0,0,0

1,0,1 
4,0,4

9,0,9

16,0,16 
25,0,25 
36,0,36

1

1,0,1

0,2,2

3,4,5

8,6,10 
15,8,17

24,10,26 
35,12,37

2

4,0,4

3,4,5 
0,8,8

5,12,13 
12,16,20 
21,20,29 
32,24,40

3

9,0,9

8,6,10

5,12,13

0,18,18

7,24,25

16,30,34 
27,36,45

4

16,0,16

15,8,17 
12,16,20

7,24,25 
0,32,32

9,40,41 
20,48,52

5

25,0,25

24,10,26

21,20,29

16,30,34

9,40,41

0,50,50

11,60,61

As this table is symmetric with respect to the diagonal, we only
need to take one half of this table as divided by the diagonal.
Let us take the upper half (m>n), and then move its content to
form a new table with parameters n and p = mn.
With this new parameter, the condition (m even, n
odd) or (m odd, n even) takes the simple form of (p
odd). The condition that m and n are coprimes, simply says
that n and p are coprimes.
Replacing m = p+n in the formulas, gives
a = p^{2}+ 2 np
b = 2 np + 2n^{2} = 2 n(p+n)
c = p^{2} + 2 np+ 2n^{2
}= p^{2}+ 2n(p+n)
thus the following table
n \ p

1

3

5

7

9

11

0

1,0,1

9,0,9

25,0,25

49,0,49

81,0,81

121,0,121

1

3,4,5 
15,8,17 
35,12,37 
63,16,65 
99,20,101 
143,24,145

2

5,12,13 
21,20,29 
45,28,53 
77,36,85 
117,44,125

165,52,173 
3

7,24,25 
27,36,45 
55,48,73 
91,60,109 
135,

187,84,205 
4

9,40,41 
33,56,65 
65,72,97 
105, 88, 137 
153,104,185 
209,120,241 
5

11,60,61 
39,80,89 
75,100,125 
119, 120, 169 
171,140, 221 
231,160,281 
6

13,84,85 
45,108,117 
85, 132, 157 
133,156, 205 
189,

253,

7

15,112,113 
51,140,149 
95, 168, 193 
147,

207,

275,

8

17,144,145 
57,176,185 
105, 208, 233 
161, 240, 289 
225,


A video on Pythagorean triples was made by Grant Sanderson (3Blue1Brown).
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