Pythagorean triples and the divisors 3, 4, and 5.

Claim 3   If $(a,b,c)$ is a solution in positive integers to the equation $a^2+b^2=c^2$ then 3 divides $a$ or $b$ , 4 divides $a$ or $b$ , and 5 divides $a$ , $b$ , or $c$ .

Proof. (The divisor 3)

Every integer squared equals 0 or 1 modulo 3. Hence, if $3\nmid ab$ then $a^2+b^2=c^2= 1+1$ =2 modulo 3. But $c^2=$ 0 or 1 modulo 3. Therefore $3 \vert ab$ . $\qedsymbol$

Proof. (The divisor 4)

Every square integer equals 0, 1, 4, or 9 modulo 16. If $4\nmid b$ then $b^2$ equals either of 1, 4, or 9 modulo 16. But none of $1+1, 1+4, 1+9, 4+4, 4+9$ , or $9+9$ equals 0, 1, 4, or 9 modulo 16. However, $c^2$ equals 0, 1, 4 or 9 modulo 16. Hence $a^2$ must equal 0 modulo 16. That is, 16 divides $a^2$ and 4 divides $a$ . $\qedsymbol$

Proof. (The divisor 5)

Every integer squared equals 0, 1, or 4 modulo 5. If $5\nmid ab$ then $a^2+b^2=$ 1+1, 1+4, or 4+4 equals 2, 0, or 3 modulo 5. But $c^2$ equals 0, 1, or 4 modulo 5. Therefore $a^2+b^2=c^2=$ 0 modulo 5 implies $5\mid c$ . $\qedsymbol$

A few generalizations

In the section entitled Primitive Pythagorean triangles where the hypotenuse is to a power the following generalizations are found for the divisors 3, 4, and 5.

Let $a^2+b^2=\left(c^k\right)^2$ where $k\in \mathbb{Z}^+$ . Let $d$ be a positive integer divisor of $k$ .

i.

If $p=4d-1$ is a prime then $p$ divides either $a$ or $b$ .

ii.

If $d=2^j$ , $j$ a non-negative integer, then $2^{j+2}$ divides either $a$ or $b$ .

iii.

If $q=4d+1$ is a prime then $q$ divides one of $a, b$ , or $c$

Set $d=1$ then $p=4(1)-1=3$ , $d=2^j=1$ implies $j=0$ implies $2^{0+2}=4$ , and $q=4(1)+1=5$ .

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