Buckets:
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| {"type":"item.completed","item":{"id":"item_0","type":"agent_message","text":"Let\n\\[\nx^2+y^2+1=kxy\n\\]\nfor some positive integer \\(k\\).\n\nIf \\(x=1\\), then \\(y\\mid y^2+2\\), so \\(y\\mid2\\). Thus \\((x,y)=(1,1)\\) or \\((1,2)\\), and both give \\(k=3\\).\n\nNow suppose \\(2\\le x\\le y\\). Define\n\\[\nz=kx-y=\\frac{x^2+1}{y}.\n\\]\nThis is a positive integer. We cannot have \\(y=x\\), since then \\(x^2\\mid1\\). Hence \\(y\\ge x+1\\), giving\n\\[\n0<z\\le\\frac{x^2+1}{x+1}\n=x-1+\\frac2{x+1}<x.\n\\]\nMoreover,\n\\[\nz^2+x^2+1=kzx,\n\\]\nbecause \\(y\\) and \\(z=kx-y\\) are the two roots of\n\\[\nt^2-kxt+x^2+1=0.\n\\]\n\nThus every solution with \\(x\\ge2\\) produces a smaller solution \\((z,x)\\) with the same \\(k\\). Repeated descent must reach a solution whose first coordinate is \\(1\\). Consequently, **every solution has \\(k=3\\)**.\n\nThe descent is reversible: from \\((z,x)\\), the original pair is uniquely\n\\[\n(x,3x-z).\n\\]\nStarting with \\((1,1)\\), this gives\n\\[\n(1,1),\\ (1,2),\\ (2,5),\\ (5,13),\\ (13,34),\\ (34,89),\\ (89,233),\\ldots\n\\]\nEvery pair satisfies the equation, since the transformation preserves it. Every solution occurs in this sequence, since its descent ends at either \\((1,1)\\) or \\((1,2)\\), and the latter is already the next pair in the sequence.\n\nAfter \\((1,1)\\), the coordinates strictly increase: if \\(a<b\\), then \\(3b-a>b\\). Therefore, once the second coordinate reaches \\(233\\), no further pair can satisfy \\(y\\le100\\).\n\nANSWER: (1, 1), (1, 2), (2, 5), (5, 13), (13, 34), (34, 89)"}} | |
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Xet Storage Details
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