Integer side, prime area?

Let the sides are $a, b, c$ , area is $P$ . There exist a theorem which shows the relationship between the sides of triangle and its area, which is $\sqrt{(s)(s-a)(s-b)(s-c)}=P \space \space(s=\frac{a+b+c}{2})$ $\frac{1}{4}\sqrt{(a+b+c)(a+b-c)(a-b+c)(-a+b+c)}=P$ $(a+b+c)(a+b-c)(a-b+c)(-a+b+c)=16P^2$ Let $\begin{aligned} a+b+c&=w\\ a+b-c&=x\\ a-b+c&=y\\ -a+b+c&=z \end{aligned}$ We can get $w=x+y+z$ , $a=\frac{w-z}{2}$ , $b=\frac{w-y}{2}$ , $c=\frac{w-x}{2}$ . Hence, $w,x,y,z$ must share same parity, because $a,b,c$ are integers. But $wxyz=16P^2$ , which means one of $w,x,y,z$ is even, so $w,x,y,z$ must be even numbers.

Let $w=2w^{'}$ , $x=2x^{'}$ ,. $y=2y^{'}$ , $z=2z^{'}$ . By ( $wxyz=16P^2$ ), and ( $w=x+y+z$ ), we can get ( $P^2=w^{'}x ^{'}y ^{'}z^{'}$ ) and ( $w^{'}=x ^{'}+y ^{'}+z^{'}$ )(obviously $w$ is the most biggest). Which means that we need to factorize $P^2$ into four integers, such that sum of the three factors equal to the biggest factor. Let $P$ is a prime number, $P^2$ can be express as ( $1\times 1\times P\times P$ ) or ( $1 \times 1 \times 1 \times P^2$ ) only. Obviously, $1+1+P>P$ , and $1+1+1=3\neq P^2$ as $3$ is not a perfect square.

Hence, there $\large does\space not$ exist triangle with integer side with prime area.