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

x,y,z are positive integers.  find all solutions of x^2 +y^2 +1=xyz.

$${x},{y},{z}\:{are}\:{positive}\:{integers}. \\ $$$${find}\:{all}\:{solutions}\:{of}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{1}={xyz}. \\ $$

Commented by MJS last updated on 24/Mar/19

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

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

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

thank you sir! i need time to come  behind it. is z=3 the only solution?

$${thank}\:{you}\:{sir}!\:{i}\:{need}\:{time}\:{to}\:{come} \\ $$$${behind}\:{it}.\:{is}\:{z}=\mathrm{3}\:{the}\:{only}\:{solution}? \\ $$

Commented by MJS last updated on 25/Mar/19

I found no other solution but I′m not sure...

$$\mathrm{I}\:\mathrm{found}\:\mathrm{no}\:\mathrm{other}\:\mathrm{solution}\:\mathrm{but}\:\mathrm{I}'\mathrm{m}\:\mathrm{not}\:\mathrm{sure}... \\ $$

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