Generating integral triangles with two equal sides How can I generate all triangles which have integral sides and area, and exactly two of its three sides are equal?
For example, a triangle with sides ${5,5,6}$ satisfies these terms.

Area of isosceles triangle with sides $a,a,b$ is $\frac{b\sqrt{4a^2-b^2}}{4}$
Now $\sqrt{4a^2-b^2}$ must be integer$=c$(say),
$\implies 4a^2-b^2=c^2\implies b^2+c^2=4a^2$
Clearly, $b,c$ are of same parity.
If $b,c$ are odd $=2C+1,2D+1$ respectively, then $b^2+c^2=(2C+1)^2+(2D+1)^2$
$=4(C^2+D^2+C+D)+2\equiv 2\pmod 4$,not multiple of $4$.
$\implies b,c$ are even
Let $b=2B,c=2C$
So, $B^2+C^2=a^2$
Area of the isosceles triangle becomes $B\sqrt{a^2-B^2}=BC$
The parametric solution of  $B^2+C^2=a^2$ is $k(p^2-q^2), 2pqk, k(p^2+q^2)$ where $p,q,k$ are any non-negative integers and $p>q$ .
So, the general solutions are $a=k(p^2+q^2), b=2B=2k(p^2-q^2),c=2C=2(2pqk)=4pqk$
or $a=k(p^2+q^2), b=2B=4pqk,c=2C=2k(p^2-q^2)$, the area being $pq(p^2-q^2)$, in either case.
If $k=1, p=2,q=1,a=2^2+1^2=5,b=2(2^2-1^2)=6$ or $b=4\cdot 2\cdot 1=8$
