Integer Isosceles Triangles

We'll show there are no solutions . Say the sides of the triangle are $a,a,b$ . Then the perimeter is $P=2a+b$ , the semiperimeter is $s=a+\frac{b}{2}$ and by Heron's formula, the area is $A=\sqrt{s(s-a)(s-a)(s-b)}$

Substituting $A=P=2s$ , $\begin{aligned} 2s&=\sqrt{s(s-a)(s-a)(s-b)} \\ 4s^2 &= s(s-a)^2 (s-b) \\ 4s &= (s-a)^2 (s-b) \\ 4\left(a+\frac{b}{2} \right) &= \left(\frac{b}{2}\right)^2 \left(a-\frac{b}{2} \right) \\ b^3-2ab^2+16b+32a&=0 \end{aligned}$

From this we see $b$ must be even; so let $b=2b'$ and divide through by $8$ to get $b'^3-ab'^2+4b'+4a =0$

At this point note we have a linear equation in the variable $a$ ; rearranging we find $a=\frac{b'\left(b'^2+4\right)}{\left(b'^2-4\right)}=b'+\frac{4}{b'+2}+\frac{4}{b'-2}$ (with some partial fraction work.)

It's simple to show that for all integer $b'>8$ , $0<\frac{4}{b'+2}+\frac{4}{b'-2}<1$ so the right hand side above cannot be an integer; it's then casework to show no other $b'$ works.