### Pythagorean array theorem

Pythagorean array theorem: each primitive Pythagorean array (a, B, c) (where a is odd and B is even) can be obtained from the following formula:

a = st\\

b = \frac{s^2-t^2}{2}\\

c = \frac{s^2 + t^2}{2}

amongs>t\geq 1Is any odd number that has no common factor.

The proof process is as follows:

Considering the interchangeability of a and B, our problem is reduced to solving the equation

A ^ 2 + B ^ 2 = C ^ 2, a is odd, B is even, a, B, C have no common factors

The tools we use are factorization and divisibility.

If (a, B, c) is a primitive Pythagorean array, it can be factorized

a^2 = c^2 – b^2 = (c – b)(c + b)

In the following examples, notice that we always assume that a is odd and B is even:

3^2 = 5^2 – 4^2 = (5 – 4)(5 + 4) = 1·9\\

15^2 = 17^2 – 8^2 = (17- 8)(17 + 8) = 9·25\\

35^2 = 37^2 – 12^2 = (37 -12)(37 + 12) = 25·49\\

33^2 = 65^2 – 56^2 = (65 – 56)(65 + 56) = 9·121

it seems thatc – bAndc + bIt’s always square.

It can be seen from the previous list,c -bAndc + bThere seems to be no common factor.

We can prove this assertion: suppose positive integersdyesc-bAndc + bThe common factor of, i.edto be divisible byc-bAndc+b。 Then D is also divisible(c + b) + (c – b) = 2cAnd(c + b) – (c – b) = 2b

So D divides 2B and 2c, but B and C have no common factor, because we assume that (a, B, c) is a primitive Pythagorean array, so d must be equal to 1 or 2. But D is also divisible（c – b)(c + b) = a^2AndaIs odd, so d must be equal to 1. In other words, dividec – bAndc + bThe number of can only be1, soc – bAndc + bThere is no common factor.

Now we knowc – bAndc + bThere is no common factor and because of(c-b)(c+b) = a^2, soc-bAndc + bThe product of is the square. Only whenc-bAndc + bIt appears only when it is square

c + b = s^2

c – b = t^2

amongs>t\geq 1Is an odd number without a common factor. By solving the above two equations

c = \frac{s^2 + t^2}{2}\\

b = \frac{s^2 + t^2}{2}\\

a = \sqrt{(c – b)(c + b)} = st

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