Gerardin gave a solution of A^4 + hB^4 = C^4 + hD^4.
(2n^2)^4 + h(n-1)^4 = (2n)^4 + h(n+1)^4, h=2n^3(n^2-1)([1].Dickson: p.648)

Ajai Choudhry[2] recently showed some new parametric solutions  of A^4 + hB^4 = C^4 + hD^4,
(h,A,B,C,D)=(p^4+3p^2+1, p^2+p+1, p-1, p^2-p+1, p+1),(2p^4-2, p^2+2p-1, p-1, p^2-2p-1, p+1),
(p^2+1, p^3+2p+1, p^2-p+1, p^3+2p-1, p^2+p+1).

We show other new parametric solutions  of A^4 + hB^4 = C^4 + hD^4.
Since there are many solutions, only a part of the solutions are shown.



Table 

   
   Table of solutions  of A^4 + hB^4 = C^4 + hD^4.
   
         h                             A       B         C        D 
    
[ (n^2+4)(n^2+2),                  n^2+n+2,  n-1,      n^2-n+2,  n+1 ]

[ 2n^4+12n^2+2,                    n^2+2n+1, n-1,      n^2-2n+1, n+1 ] 

[ 2(n^2+9)(n^2+3),                 n^2+2n+3, n-1,      n^2-2n+3, n+1 ]

[ 3(n^2+4)(n^2-2),                 n^2+3n-2, n-1,      n^2-3n-2, n+1 ]

[ 3n^4+33n^2+3,                    n^2+3n+1, n-1,      n^2-3n+1, n+1 ]

[ n^6+2n^4+n^2+1,                  n^3+n+1,  n-1,      n^3+n-1,  n+1 ]

[ 2n^6+4n^4+2n^2+8,                n^3+n+2,  n-1,      n^3+n-2,  n+1 ]

[ 3n^6+6n^4+3n^2+27,               n^3+n+3,  n-1,      n^3+n-3,  n+1 ]

[ (n^2+2)(n^4+3n^2+1),             n^3+2n+1, n-1,      n^3+2n-1, n+1 ]

[ (2n+1)(n^6+2n^4+5n^2+4n+1),      n^3+3n+1, n-1,      n^3-n-1,  n+1 ]

[ 2(n+1)(n^6+2n^4+5n^2+8n+4),      n^3+3n+2, n-1,      n^3-n-2,  n+1 ]

[ 2(n+1)(n^2-n+2)(n^4+n^2+4n+4),   n^3+3n+2, n-1,      n^3-n+2,  n+1 ]

[ 2(n^2+4)(n^2+3)(n^2+1),          n^3+3n+2, n-1,      n^3+3n-2, n+1 ]

[ (n^2+3)(n^2+1)^2,                n^3+3n+2, n-2,      n^3+3n-2, n+2 ]

[ (2n+3)(n^6+2n^4+5n^2+12n+9),     n^3+3+3n, n-1,      n^3-n-3,  n+1 ]


Reference

[1] L.E. Dickson, History of theory of numbers, vol 2.

[2] Ajai Choudhry, A Note on the Quartic Diophantine Equation A^4+ hB^4= C^4+ hD^4.
     https://arxiv.org/pdf/1608.06220v1