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) On multiplying by n^4, leads to a solution, (2n^3)^4 + h(n-1)^4 = (2n^2)^4 + h(n+1)^4, h=2n^7(n^2-1). Similarly, on multiplying by 8n(n^2-1)^3, leads to a solution, (2n(n-1)(n^2-1))^4 + h(2n^2)^4 = (2n(n+1)(n^2-1))^4 + h(2n)^4, h=8n(n^2-1)^3 We show three new solutions.Solutions1. Use a known solution.Assume p^4 + hq^4 = ha^4 + r^4. Let A = ax+p, B = x+q, C = ax+r, D = x+s,.......................................(1) A^4 + hB^4 - C^4 - hD^4 =(-4ra^3+4hq-4ha+4pa^3)x^3 +(-6ha^2-6r^2a^2+6hq^2+6p^2a^2)x^2 +(-4ha^3+4p^3a-4r^3a+4hq^3)x +p^4+hq^4-ha^4-r^4......................................(2) Equating to zero the coefficient of x in (2), then we obtain a = -h(-s^3+q^3)/(p^3-r^3) x =-3/2(-q^4ar^4h-q^3a^2r^4h+q^2r^4ha^3+qr^4ha^4+pha^5r^3-p^3rha^5-q^5phr^3+q^5p^3rh+q^4ap^4h+q^3a^2p^4h-q^2p^4ha^3-qp^4ha^4-q^4aphr^3+q^4ap^3rh -q^3a^2phr^3+q^3a^2p^3rh+q^2pha^3r^3-q^2p^3rha^3+qpha^4r^3-qp^3rha^4+q^5p^4h+p^7ra+p^7rq+p^6r^2a+p^6r^2q-2p^5qr^3-2p^5ar^3-2p^4r^4q-2p^4r^4a -2p^3r^5q-2p^3r^5a+p^2qr^6+p^2ar^6+pr^7q+pr^7a+r^8a+p^8a+p^8q+r^8q+r^4ha^5-q^5r^4h-p^4ha^5) /(-h^2q^8-h^2aq^7-h^2a^2q^6+2h^2a^3q^5+2h^2a^4q^4+2h^2a^5q^3-h^2a^6q^2-h^2a^7q-2p^4r^4+p^2r^6+p^8+p^7r-2p^3r^5+p^6r^2-2p^5r^3+pr^7+r^8-h^2a^8): Substitute p = 2n^2, q = n-1, r = 2n, s = n+1, and h = 2n^3(n^2-1) to (1), then we obtain a new solution. A = (13n^11-41n^10+19n^9+25n^8-62n^7+70n^6-18n^5+10n^4-15n^3+3n^2-n-3)n B = (n^4-4n^3+4n^2-1)(5n^7-13n^6-33n^5-29n^4-9n^3+n^2+5n+1) C = (9n^11-13n^10-73n^9+125n^8+26n^7-34n^6-58n^5-30n^4+29n^3+15n^2+3n+1)n D = (n^4-4n^3+4n^2-1)(5n^7+9n^6+23n^5+21n^4+31n^3+11n^2+5n-1) Similarly, by using this new solution as a known solution, we obtain more new solution.2. Find some solutions of quadratic equation about h.A^4 + hB^4 - C^4 - hD^4........................................................(3) Let A = px+a, B = qx-1, C = px-a, D = qx+1, then (3) becomes to (4). (8ap^3-8hq^3)x^3+(-8hq+8a^3p)x.................................................(4) To obtain a solution of (4), we have to find the rational solution of (5). v^2=-64a^4p^4+64ahqp^3+64hq^3a^3p-64h^2q^4.....................................(5) Let consider (5) is a quadratic equation for h. We found two solutions of (5) as follows.2-1. h = p^3a/qSubstitute h = p^3a/q to (5), then v^2 = -64p^4a^2(p-a)(p+a)(q-1)(q+1). Let a = m^2+n^2, p = m^2-n^2, q = (m^2+n^2)/(2mn), then h = 2(m^2-n^2)^3mn, x = 4m^2n^2/(-2m^2n^2+n^4+m^4) We obtain a solution.(degree 6) A = (n-m)(n+m)(n^4-4m^2n^2-m^4) B = 2m^3n+2n^3m+2m^2n^2-n^4-m^4 C = (n-m)(n+m)(n^4+4m^2n^2-m^4) D = 2m^3n+2n^3m-2m^2n^2+n^4+m^4 h = 2mn(m^2-n^2)^3 By using this new solution as a known solution, we obtain more new solutions.2-2. h = pa/qSimilarly, (5) becomes to v^2 = -64a^2p^2(a-1)(a+1)(p-q)(p+q). Let a = (m^2+n^2)/(2mn), q = m^2+n^2, p = m^2-n^2, then h = 8(m^2-n^2)m^3n^3, x = -1/4(-m^2+n^2)/(m^2n^2) We obtain a solution.(degree 6) A = 2mn(m^4-2m^2n^2+n^4+2m^3n+2mn^3) B = m^4-n^4-4m^2n^2 C = 2mn(m^4-2m^2n^2+n^4-2m^3n-2mn^3)mn D = m^4-n^4+4m^2n^2 h = 8m^3n^3(m^2-n^2)[1].L.E. Dickson:History of theory of numbers, vol 2.Reference