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On the Diophantine equations $$z^2=f(x)^2 \pm f(y)^2$$ z 2 = f ( x ) 2 ± f ( y ) 2 involving quartic polynomials

Crossref DOI link: https://doi.org/10.1007/s10998-018-0259-7

Published Online: 2018-10-12

Published Print: 2019-09

Update policy: https://doi.org/10.1007/springer_crossmark_policy

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