On $X_1^4+4X_2^4=X_3^8+4X_4^8$ and $Y_1^4=Y_2^4+Y_3^4+4Y_4^4$
Communications in Mathematics, Tome 23 (2015) no. 2, pp. 113-117.

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The two related Diophantine equations: $X_1^4+4X_2^4=X_3^8+4X_4^8$ and $Y_1^4=Y_2^4+Y_3^4+4Y_4^4$, have infinitely many nontrivial, primitive integral solutions. We give two parametric solutions, one for each of these equations.
Classification : 11D41, 11D72
Keywords: Diophantine equation $A^4+nB^4=C^2$; Diophantine equation $A^4-nB^4=C^2$; Diophantine equation $X_1^4+4X_2^4=X_3^8+4X_4^8$; Diophantine equation $Y_1^4=Y_2^4+Y_3^4+4Y_4^4$
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     title = {On $X_1^4+4X_2^4=X_3^8+4X_4^8$ and $Y_1^4=Y_2^4+Y_3^4+4Y_4^4$},
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Jena, Susil Kumar. On $X_1^4+4X_2^4=X_3^8+4X_4^8$ and $Y_1^4=Y_2^4+Y_3^4+4Y_4^4$. Communications in Mathematics, Tome 23 (2015) no. 2, pp. 113-117. http://geodesic.mathdoc.fr/item/COMIM_2015__23_2_a1/