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on equation $2000x^2 + y^2 = 2001z^2.$

Source: 12-th Hungary-Israel Binational Mathematical Competition 2001

April 11, 2007
Diophantine equationnumber theory proposednumber theory

Problem Statement

Find positive integers x,y,zx, y, z such that x>z>1999⋅2000⋅2001>yx > z > 1999 \cdot 2000 \cdot 2001 > y and 2000x2+y2=2001z2.2000x^{2}+y^{2}= 2001z^{2}.
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