If $A^2+B^2=C^2$ with $A$ odd, $A,B,C$ coprime and $A<B<C$, is $B+C$ a square?

Question

I was looking through Pythagorean triplets and noticed something:

Take $3$ numbers $A,B,C$ such that:

1. $A^2+ B^2= C^2$
2. $A<B<C$
3. $A,B$ and $C$ have no common factors
4. $A$ is odd

Prove that $B+C$ is always a perfect square

For example $3^2+ 4^2= 5^2$ and $4+5=9= 3^2$

$5^2+ 12^2= 13^2$ and $12+13=25= 5^2$

$7^2+ 24^2= 25^2$ and $24+25=49= 7^2$

$7^2+ 24^2= 25^2$ and $24+25=49= 7^2$

$33^2+ 56^2= 65^2$ and $56+65=121= 11^2$

$65^2+ 72^2= 97^2 $ and $72+97=169= 13^2$

$19^2+ 180^2= 181^2 $ and $180+181=361= 19^2$

I noticed this, can anyone prove it ?

Answer 1

Assuming that $A$, $B$ and $C$ are positive, it follows from the conditions that $$A^2=C^2-B^2=(C+B)(C-B).$$ The two factors on the right hand side are coprime, because $A$, $B$ and $C$ have no common factor and $A$ is odd. It follows that both factors are perfect squares because $B$ and $C$ are positive.

Answer 2

Disregard the part about $\space A<B\space $ because $\space A>B\space $ for half of all triples such as $(15,8,17).\quad $ Now to the square part. Both the sum and difference are squares, given Euclid's formula: $$A=m^2-k^2\quad B=2mk\quad C=m^2+k^2$$

$$C\pm B= (m^2+k^2) \pm 2mk = (m\pm k)^2$$