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Proposition. Let [math]a, b, c \in \mathbb{N\setminus \{0\}}[/math]. For all odd primes [math]a[/math] there exist [math]b, c[/math] such that [math]a^2 + b^2 = c^2[/math]. Namely [eqn]a\ \text{odd prime} \implies b = \frac{a^2 - 1}{2} \land c = b + 1.[/eqn]

Proof. Subtract [math]b^2[/math] to yield [math]a^2 = c^2 - b^2[/math]. Factorize into [eqn]a^2 = (c + b)(c - b).\quad(1)[/eqn]
Note that [math]c + b[/math] and [math]c - b[/math] are integers because theyre sums of integers.
Let [math]a[/math] be odd prime. We know that the only divisors of a prime squared are [math]\{1, a, a^2\}[/math]. Thus there are only 3 cases for eq 1.
(I) [math](c + b = 1 \land c - b = a^2) \lor[/math]
(II) [math](c + b = a \land c - b = a) \lor[/math]
(III) [math](c + b = a^2 \land c - b = 1).[/math]
Case I. [math]c + b = 1[/math] is impossible if [math]b, c \in \mathbb{N} \setminus \{0\}[/math].
Case II. This implies that [math]b = -b[/math] which is impossible for nonzero reals (may i remind you that [math]b \in \mathbb N\setminus \{0\}[/math]).
So case III holds. And we know it MUST hold (means "no solution" is impossible) because of the following.
From (III) derive [math]c = 1 + b[/math] and substitute it to get [math](1 + b) + b = a^2 \iff b = \frac{a^2 - 1}{2}[/math].
Write [math]a = 2k + 1,\quad \forall k \in \mathbb N \setminus \{0\}[/math] (definition of odd numbers). Then [eqn]b = \frac{(4k^2 + 4k + 1) - 1}{2} = 2k^2 + 2k \in \mathbb{N} \land c = b + 1 \in \mathbb{N}.[/eqn]
Note that this is not the case for a = 2. In fact there are no integer solutions to [math]2^2 + b^2 = c^2[/math].
Therefore ALL odd primes [math]a[/math] generate a natural [math]b[/math] and [math]c[/math].
Q.E.D.

Example. If a = 13 then b should be (13^2 - 1) / 2 = 84 and c should be 84 + 1 = 85. This is true 13^2 + 84^2 = 85^2.