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Question Number 56800 by mr W last updated on 24/Mar/19

$$x,y,zarepositiveintegers.$$$$findallsolutionsof{x}^2+y^2+1=xyz.$$

Commented by MJS last updated on 24/Mar/19

$$\mathrm{with}z=3\mathrm{we}\mathrm{get}\mathrm{pairs}\mathrm{of}\mathrm{Fibonacci}\mathrm{numbers}$$$$\mathrm{for}(x,y)\mathrm{or}(y,x)\mathrm{with}\mathrm{each}{2}^{\mathrm{nd}}\mathrm{one}\mathrm{missing}$$$$(1,2)$$$$(2,5)$$$$(5,13)$$$$(13,34)$$$$(34,89)$$$$...$$

Commented by mr W last updated on 25/Mar/19

$$thankyousir!ineedtimetocome$$$$behindit.isz=3theonlysolution?$$

Commented by MJS last updated on 25/Mar/19

$$\mathrm{I}\mathrm{found}\mathrm{no}\mathrm{other}\mathrm{solution}\mathrm{but}{\mathrm{I}}^{\prime }\mathrm{m}\mathrm{not}\mathrm{sure}...$$

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