Pythagoras theorem (Pythagorean theorem) indicates that the sum of the squares of the lengths of the two right angles of any right triangle is equal to the square of the length of the hypotenuse. It is expressed by formula

a^2 + b^2 = c^2

The first question is whether there are infinitely many Pythagorean arrays that satisfy the equationa^2 + b^2 = c^2The natural number triples of (a, B, c). The answer is yes. If we take the Pythagorean array (a, B, c) and multiply it by an integer D, we get a new Pythagorean array (DA, DB, DC). This is true because

(da)^2 + (db)^2 = d^2(a^2 + b^2) = d^2c^2 = (dc)^2

### Original Pythagorean array (PPT)

The primitive Pythagorean array is a triple (a, B, c), where a, B, C have no common factors and satisfy the

a^2 + b^2 = c^2

Here are some primitive Pythagorean arrays obtained:

(3, 4, 5)\ \ (5, 12, 13)\ \ (8, 15, 17)\ \ (7, 24, 25)\\

(20, 21, 29)\ \ (9, 40, 41)\ \ (12, 35, 37)\ \ (11, 60，61)\\

(28, 45, 53)\ \ (33, 56, 65)\ \ (16, 63, 65)

It is easy to draw some conclusions from this short table. For example, it seems that a and B have different parity and C is always odd.

It is proved that: first of all, if a and B are even, then C is also even, which means that a, B and C have a common factor of 2, so the triple ancestor is not primitive. Secondly, if a and B are odd numbers, then C must be even. So there are integers x, y, Z such that

a = 2x + 1, b = 2y + 1, c = 2z

Bring it into the equationa^2 + b^2 = c^2have to

(2x + 1)^2 + (2y + 1)^2 = (2z)^2\\

4x^2 + 4x + 4y^2 + 4y + 2 = 4z^2

Divide both sides by two

2x^2 + 2x + 2y^2 + 2y + 1 = 2z^2

The last equation says that an odd number is equal to an even number, which is impossible, so a and B cannot both be odd. Because we have proved that they can’t all be even, and they can’t all be odd, so their parity is different. Then by the equationa^2 + b^2 = c^2C is odd.

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