id	author	title	date	pages	extension	mime	words	sentence	flesch	summary	cache	txt
ejpam-4940	Sibihi, Alanoud	Solutions of Some Quadratic Diophantine Equations	2023	10	.pdf	application/pdf	3117	158	80	Then u2k − P (t)v2k = (a0y(k−1)l−1xl−1 + x(k−1)l−1xl−1 + y(k−1)l−1xl−2) 2 −P (t)(a0y(k−1)l−1yl−1 + x(k−1)l−1yl−1 + y(k−1)l−1yl−2) 2 = (u1uk−1 + (a0u1 + α)vk−1) 2 − P (t)(v1uk−1 + (a0v1 + β)vk−1) 2 = u21u 2 k−1 + 2u1(a0u1 + α)uk−1vk−1 + (a0u1 + α)2v2k−1 −P (t)v21u 2 k−1 − 2P (t)(a0v1 + β)v1uk−1vk−1 − P (t)(a0v1 + β)2v2k−1 = (u21 − P (t)v21)u 2 k−1 − [ (P (t)(a0v1 + β)2 − (a0u1 + α)2 ] v2k−1 +2 [u1(a0u1 + α)− P (t)v1(a0v1 + β)]uk−1vk−1 Using the above lemma, we have (P (t)(a0v1 + β)2 − (a0u1 + α)2 = P (t)u21 − P (t)2v21 = P (t)(u21 − P (t)v21) = P (t) and u1(a0u1 + α)− P (t)v1(a0v1 + β) In [2], Tekcan consider the number of integer solutions of Diophantine equation E : x2 − (t2 − t)y2 − (4t − 2)x + (4t2 − 4t)y = 0 over Z, where t ≥ 2.	cache/ejpam-4940.pdf	txt/ejpam-4940.txt