Number of Pythagorean triples between $10<a+b-c<18$

Question

A competition problem asks:

How many Pythagorean triples $(a,b,c)$ ($c$ being the hypotenuse) exist such that:

$$10<a+b-c<18?$$

Answer 1

Let $gcd(a,b,c)=1$ and $a$ be an even natural number.

Thus, there are naturals $m$ and $n$ with different parity such that $m>n$, $gcd(m,n)=1$ for which $a=2mn$, $b=m^2-n^2$ and $c=m^2+n^2$.

Thus,

$$10<2mn+m^2-n^2-m^2-n^2<12$$ or $$5<n(m-n)<9$$ or $$6\leq n(m-n)\leq8$$ and the rest is a finite number of cases and systems.

After all this we need to check $d(a,b,c)$, where $d=gcd(a,b,c)$

and we remember that we can change $a$ with $b$.