$ S[a,b],R[b,c]$ So we has: $ 2(a+b)=cd$ $ 2(c+d)=ab$ $ \Rightarrow 4(a+b)(c+d)=abcd$

... and the solutions of the last equality?

Peter wrote: ... and the solutions of the last equality? WLOG $ a\ge b$ and $ c\ge d$ and $ b\ge d$ $ abcd = 4(a + b)(c + d)\le 4(a + a)(c + c) = 16ac\Rightarrow bd\le16 \Rightarrow d^2\le bd\le16 \Rightarrow d\le 4$ $ \\ 2(c + d) = ab\Rightarrow 2c = ab - 2d$ $ 2(a + b) = cd \Rightarrow 4(a + b) = 2cd \Rightarrow 4(a + b) = (ab - 2d)d \Rightarrow abd^2 - 4ad - 4bd - 2d^3 = 0 \Rightarrow \\ (ad - 4)(bd - 4) = 2d^3 + 16$ and obviously we can put $ d = 1,2,3,4$ in the last equality,and find solutions... $ d = 1$: $ (a - 4)(b - 4) = 18$ $ \Rightarrow b = 5,a = 22,c = 54$ or $ b = 6,a = 13,c = 38$ or $ b = 7,a = 10,c = 34$ $ d = 2$: $ (2a - 4)(2b - 4) = 32\Rightarrow (a - 2)(b - 2) = 8$ $ \Rightarrow b = 3,a =10 ,c =13$ or $ b = 4,a =6,c = 10$ $ d = 3$: $ (3a - 4)(3b - 4) = 70$ $ \Rightarrow b = 3,a = 6,c = 6$ $ d = 4$: $ (4a - 4)(4b - 4) = 144\Rightarrow (a - 1)(b - 1) = 9$ $ \Rightarrow b = 4,a = 4,c = 4$ therefore all solutions are: $ \{\{a,b\},\{c,d\}\} = \{\{22,5\},\{54,1\}\},\{\{13,6\},\{38,1\}\},\{\{7,10\},\{34,1\}\} \\ ,\{\{10,3\},\{13,2\}\},\{\{6,4\},\{10,2\}\},\{\{6,3\},\{6,3\}\},\{\{4,4\},\{4,4\}\}$