We need to find all cyclic quadrilaterals (or formulas that gives its sides), which have integer sides $a,b,c,d$. The constrain is that its area must be an integer multiple of its perimeter. We can find its area by Brahmagupta's formula: $$s=(a+b+c+d)/2 \implies A=[(s-a)(s-b)(s-c)(s-d)]^{1/2}$$ And as you know its perimeter is $P=a+b+c+d=2s$. For all integer $k$, we need to find $(a,b,c,d)$ integer quadruples that satisfy $A=kP$.
On
I became entangled in the problem and decided another equation. It is necessary to solve this equation:
$$S_4=\frac{1}{4}\sqrt{(a+b+c-d)(a+b-c+d)(a-b+c+d)(b-a+c+d)}$$
Solutions will be:
$$a=(pt+ks)(ts-pk)n$$
$$b=(pt+ks)(ts-pk)j$$
$$c=tk(p^2+s^2)j-ps(t^2+k^2)n$$
$$d=ps(t^2+k^2)j-tk(p^2+s^2)n$$
$p,t,k,s,j,n - $ integers asked us. Then the area of the quadrilateral are equal.
$$S_4=pstk(ts-pk)(pt+sk)(j-n)(j+n)$$
And its perimeter is equal to:
$$P_4=2(pt+sk)(tsj-pkn)$$
To solve the equation:
$$S_4=FP_4$$
When the number of $F$ is set for the problem. Come to the need to solve the following equation:
$$pstk(ts-pk)(j-n)(j+n)=2F(tsj-pkn)$$
Left to think. What is the easiest way to solve this equation.
To begin, write the formula of the solution of the following equation:
$$S_4=\sqrt{(a+b+c)(a+b+d)(a+c+d)(b+c+d)}$$
Formulas of the solutions can be written.
$$a=((t^2+k^2)s-tkp)pn$$
$$b=((t^2+k^2)s-tkp)pj$$
$$c=(tkp^2-t^2ps+tks^2)n+(tkp^2-(t^2-k^2)ps-tks^2)j$$
$$d=(tkp^2+(t^2-k^2)ps-tks^2)n+(tkp^2-k^2ps+tks^2)j$$
The square will be equal to:
$$S_4=pstk((2pk-ts)j+(pk+ts)n)((pt+sk)j+(2pt-sk)n)$$
And the perimeter.
$$P_4=a+b+c+d=p((ktp+sk^2)j+t(pk+ts)n)$$
$p,k,t,s,j,n $ - Integers asked us.
And now about the idea of solving the equation:
$$S_4=FP_4$$
$F$ - the integer specified by the problem statement. To do this, put this number on the multiplier. And we will solve the linear equation in the unknown $j,n$ .
$$st((2pk-ts)j+(pk+ts)n)=F$$
The number $p,k$ determine from the equation.
$$ptk=s(t^2-k^2)$$