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    $\begingroup$ Another solution is $1+3870^4+11954^4=6953^4+11632^4$. $\endgroup$ Commented Feb 8, 2025 at 20:09
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    $\begingroup$ Somewhat related are oeis.org/A004831 (Numbers that are the sum of at most 2 nonzero 4th powers. See especially the comment) and oeis.org/A336536 (Numbers n that can be written as both the sum of two nonzero fourth powers and the sum of three nonzero fourth powers.) $\endgroup$ Commented Feb 8, 2025 at 21:17
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    $\begingroup$ There may be something in Andrzej Nowicki, Cubes, Fourth Powers and Higher Powers, available at academia.edu/30783812/Cubes_Fourth_Powers_and_Higher_Powers where 5.2.5. says, Kolejne liczby i sumy dwóch bikwadratów: $3502321 = 25^4 + 42^4$, $3502322 = 17^4 + 43^4$. The Polish translates as Consecutive numbers and sums of two biquadrates. $\endgroup$ Commented Feb 8, 2025 at 21:33
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    $\begingroup$ After Reading a References given by @GerryMyerson in Comments, 1st solution is already exists in oeis and also in Andrzej Nowicki, Cubes, Fourth Powers and Higher Powers. Thanks for References. $\endgroup$ Commented Feb 9, 2025 at 20:07
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    $\begingroup$ @David , It works only when factors = 0. Why you think that it has a polynomial parametrization with degree 2 ? $\endgroup$ Commented Feb 17, 2025 at 9:43