Let $ \Delta ABC$ be a triangle with sidelengths $ AB=c$, $ BC=a$, $ CA=b$ , semiperimeter $ s=\frac{1}{2}( a+b+c )$ and area $ A$. From the Heron formula $ A =\sqrt{s(s-a) (s-b ) ( s-c ) }$. We know that $ s=2p$ and the area $ A$ is an integer number . Thus $ 1\leq a,b,c\leq 2p-1$, $ {s(s-a) (s-b ) ( s-c ) }$ needs to be a perfect square. Consequence : there exists such a side $ m$ ($ m= a,b,c$ ) of $ \Delta ABC$ that $ p\mid (s-m )$. We can choose $ m=a$ with out lost of generality. Then $ s-a= p$, since $ 1\leq (s- a) \leq 2p-1$, then $ a=p$. From the other side $ (s-b ) + (s-c )= a$, and since $ (s-b) (s-c ) = \frac {A^2}{s (s-a ) }$ then $ (s-b )(s-c ) = 2 k^2$, where k is natural number. It is evident that $ ((s-b ),(s-c )) =1$ . Because otherwise there would be a number $ h$, such that $ h\mid (s-b)$ and $ h\mid (s-c )$ and so $ h\mid (s-b) + (s-c )$ that is $ h\mid a$, therefore $ h\mid p$ and that implies that $ h=1$ or $ h=p$ .If $ h=p$ then $ b=c=p$ then $ 2s= a+b+c =3a=3p$ , hence there is a contradiction. We know (Sierpinski) that there exists a unique representation of prime p in the form $ p=tx^2 + y^2$ with $ t,w,x,y$ natural numbers and $ (x,y)=1$. So we take $ p=2x^2 +y^2$. That means Suppose $ (s-b)= 2x^2$ and $ (s-c)= y^2$ with $ ( x,y)=1$, ($ k= x^2y^2$), then $ (s-b )+(s-c) =a= 2x^2 +y^2=p$. Hence, $ \Delta ABC$ has the sides $ a=p$, $ b= 2p-2x^2$ and $ c= 2p-y^2$ .Thus the perimeter $ 2s= p +(2p-2x^2) + (2p-y^2)= 4p,$ and the area $ A^2= s (s-p )(s-(2p-2x^2))(s-y^2 )= 4p^2 x^2y^2$ Next, it is possible if and only if $ p = 1 (mod8)$ or $ p=3 (mod8)$. That is when 1° case $ 2x^2 +y^2 = 1 (mod8)$ or 2° case $ 2x^2 +y^2 =3 (mod8)$. Example case 1° Taking $ x=4$ and $ y=3$ $ x^2=0(mod8)$ and $ y^2=1(mod8)$ we have $ p=41$ ,$ {s(s-a) (s-b ) ( s-c )=2*41*2*4*9}$, which is a perfect square. case 2° For example let $ x^2=1(mod8)$ and $ y^2=1(mod8)$. Taking $ x=3$ , $ y=5$, we have $ p=43$, then $ a=43$, $ b=68$,$ c=61$, $ {s(s-a) (s-b ) ( s-c )= 1664100}$ which is a perfect square.

Sorry, small mistake : Quote: s(s-a) (s-b ) ( s-c )=2*41*2*4*9 I meant of course: $ s(s-a) (s-b ) ( s-c )= 4\cdot 41^2 \cdot 16 \cdot 9$ and then the triangle $ \Delta ABC$ has $ a=41$ , $ b= 50$ , $ c=73$ .