Relating Perimeter and Area

First, clear denominators to get $P+4=AP-P-4A+4$

Rearranging gives $AP-2P-4A=0$ , and using Simon's Favorite Factoring trick we obtain $(A-2)(P-4)=8$ . Since the rectangle has integer perimeter and area, then $A-2$ and $P-4$ are integers, so we look at the positive factors of $8$ :

$8=1\times 8, 2\times 4, 4\times 2, 8\times 1$

Thus, $(A-2,P-4)=(1,8), (2,4), (4,2), (8,1)$ so $(A,P)=(3, 12), (4, 8),(6, 6),(10,5)$ .

We may be tempted to say the answer is $12+8+6+5=31$ , but we must first confirm that all of these rectangles exist.

So let's let $x,y$ be the side lengths of the rectangle. We have that $xy=A$ , and $2x+2y=P$ .

We substitute $(A,P)=(3, 12), (4, 8),(6, 6),(10,5)$ in and solve the 4 different system of equations to find that $(A,P)=(3, 12), (4, 8)$ yield real values for the side lengths, and $(A,P)=(6, 6),(10,5)$ don't.

So the answer is just $12+8=\boxed{20}$ .

First,

$\frac { P+4 }{ P-4 } =A-1\\ \hookrightarrow P+4=AP-P-4A+4\\ \quad \quad 2P-AP+4A=0\\ \quad \quad (P-4)(-A+2)=0-8\\ \quad \quad (P-4)(A-2)=8$ .

Now, we know that $8=1x8=2x4=4x2=8x1$ . Them, the solutions for $(P,A)$ are $(5,10)\\ (6,6)\\ (8,4)\\ (12,3)$

But, $(3,12) and (4,8)$ yield real values for the side lengths.

So the answer is $12+8=20$ .

Sorry for not formatting.

(P +4)(P - 4) = Integer.

Let P - 4 = x.

(x+8)/x should be integer. So x = 1,2,4,8. That means Perimeter = 5,6,8,12.

Now checking these values, by solving for Area for each given perimeter.

For P =5, area = 10.

L + B = 5/2; LB = 10.

(L+B)^2 = 25/4 = L^2 + B^2 + 20. This gives the sum of 2 squares as negative, so is not possible. Same with P = 6.

However, P = 8 and P = 12 satisfy as there will be possible values of L and B for their cases.

(P - 4)(A - 2) = 8

(P, A) of (5, 10), (6, 6), (8, 4) and (12, 3) are initial possibilities.

But P = 2 (a + b) and A = a b where

x^2 - 2.5 x + 10 = 0 => Real x not valid;

x^2 - 3 x + 6 = 0 => Real x not valid;

x^2 - 4 x + 4 = 0 => (x - 2)^2 = 0 => a = 2 and b = 2;

x^2 - 6 x + 3 = 0 => (x - 3)^2 = 6 => a = 3 + sqrt(6) and b = 3 - sqrt(6).

Reduced to only (8, 4) and (12, 3) as only possible answers where a square is also inclusive as a rectangle. Therefore, 8 + 12 = 20.