Having read the Wikipedia article "Pythagorean Triple", I came across the "Elementary properties of primitive Pythagorean triples" section which described many conditions for primitive triples, namely:
Exactly one of a, b is odd; c is odd.
At most one of a,b,c is a square
Exactly one of a,b is divisible by 3
Exactly one of a,b is divisible by 4
Exactly one of a,b,c is divisible by 5
I would like to know if there are proofs that exist for these statements. If so, would anyone be kind enough to (mathematically) explain to me why these statements are true?
With regard to factors of $2$ (odd/even), $3,4,5$, it should be clear that at most one of $a,b,c$ can have each of those as a factor, since if two of them had a common factor, the third would also have to have that factor and the triple would not be primitive.
Regarding $2$: If $c$ is even, then both $a,b$ must be odd. In that case, $a\equiv \pm 1 \mod{4}$ and $b\equiv \pm 1 \mod{4}$. Hence $a^2\equiv 1 \mod{4}$ and $b^2\equiv 1\mod{4}$, so $c^2\equiv 2\mod{4}$. Even numbers that are congruent to $2$ $\mod{4}$ (such as $2,6,10,14,18$, etc.) are simply twice an odd number. They contain only one factor of $2$, and hence cannot be perfect squares. So $c$ cannot be even, and must be odd. This means that both $a,b$ cannot be odd (or the sum of their squares would be even), so one must be even.
Regarding $3$: Any number $\mod{3}$ must be congruent to one of $-1,0,1$. So the square of any number $\mod{3}$ must be congruent to either $0,1$. The only way you can make a sum involving three squares $\mod{3}$ is $1+0=1$. So one of $a,b$ must be divisible by $3$.
Regarding $4$: This result is most easily understood from considering the generating formula for Pythagorean triples. All primitive Pythagorean triples can be generated from the relationships $a=m^2-n^2, b=2mn, c=m^2+n^2$ where $m>n$ are integers. It is apparent that $b$ is the even member of the triple, and for $a,c$ to be odd, it is apparent that $m,n$ must have different parity. In other words, one of $m,n$ must be even. Hence $2mn$ must be divisible by $4$.
Regarding $5$: Any number $\mod{5}$ must be congruent to one of $0,\pm 1,\pm 2$. So the square of any number $\mod{5}$ must be congruent to one of $0,\pm 1$. You can't make a sum involving three such numbers which omits $0$, that is, consists of only $1$ and $-1$. So one of the numbers must be divisible by $5$.
The final fact (at most one of $a,b,c$ can be a square) is demonstrated in the proof of Fermat's Last Theorem ($x^n+y^n=z^n$ has no solutions for $n>2$) for the specific exponent $n=4$. Beyond the scope of this answer.