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.
 A: Heron's formula says the area of a triangle with sides $a, b, c$ is
$$Area = \sqrt{s(s-a)(s-b)(s-c)}$$
where $s$ is the semiperimeter, $s = \frac{a+b+c}{2}$.
Now, your assumption is two sides are equal, so $a, b, b$.  The area is now
$$Area = \sqrt{s(s-a)(s-b)^2} = (s-b)\sqrt{s(s-a)}$$
Now $s-b$ might not be an integer, if $s$ is not, but it will at worst be a fraction of the form $\frac{z}{2}$.  I will deal more with this later.
For now, we want $\sqrt{s(s-a)}$ to be an integer, so $s(s-a)$ must be a perfect square.  Now, $s = \frac{a + 2b}{2} = \frac{a}{2} + b$ so this simplifies to wanting
$$s(s-a) = (\frac{a}{2} + b)(\frac{a}{2} + b - a) = (b + \frac{a}{2})(b - \frac{a}{2}) = b^2 - \frac{a^2}{4}$$
a perfect square, so let's say it equals $c^2$.  Thus, we want all integer solutions to
$$c^2 + \left(\frac{a}{2}\right)^2 = b^2$$
If we happen to get a solution such that $(s-b)\sqrt{s(s-a)}$ is not an integer, then multiply all side lengths by 2 to get a similar triangle with $s$ and thus $s-b$ an integer.
Other than that, the problem is reduced to finding all solutions to this equation, which is a well known problem with a well known solution.
A: 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 $b=4\cdot 2\cdot 1=8$
