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How to find the smallest positive number $x$ such that $2011x^{2}+1$ is a square number

$$ 2011x^{2}+1=y^{2} $$

$x,y$ are positive numbers

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Try using continued fractions. –  dinoboy Jan 24 '13 at 19:35
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According to Maple, $510732021116138713675018566232201605320997$ –  Robert Israel Jan 24 '13 at 19:37
    
It seems the question is not What is the smallest positive number etc. but How to find it. –  Did Jan 24 '13 at 20:02
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2 Answers 2

Dario Alpern's solver shows

$X_0 = 0\\ Y_0 = 1$

$X_{n+1} = P X_n + Q Y_n \\ Y_{n+1} = R X_n + S Y_n \\ \\ P = 22 903355 954053 525066 202335 319378 237605 968890 (44 \text{ digits})\\ Q = 510732 021116 138713 675018 566232 201605 320997 (42 \text{ digits})\\ R = 1027 082094 464554 953200 462336 692957 428300 524967 (46 \text{ digits})\\ S = 22 903355 954053 525066 202335 319378 237605 968890 (44 \text{ digits})\\$

It will give you a step by step solution

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a=0;b=1;c=19742928095;d=441354968;e=883151290968;f=19742928095 ;l=10;z=vector(l);y=vector(l);z[1]=a;y[1]=b;for(n=2,l,z[n]=c*z[n-1]+d*y[n-1];y[n]=e*z [n-1]+f*y[n-1];print([z[n],y[n]]))

(Pari gp)

[441354968, 19742928095] [17427278795190051920, 779566419528680658049] [688131024289910853831312029832, 30781847532062692044137637045215] [27171442664988816842285870050168151808160, 1215447604913533872118296302514247336972801] [1072887677544578769792061644645677816761262916480568, 47992989334095735859126116621354243081754352590442975] [42363888543548329618109290861429128960866113212881520825707760, 1895044274974308089824464492379048791500336482990271031908792449] [1672774410679737895614308190979727225255931422523452909015625403410553832, 74827445715318345179514414300506375814316454587342868891991725485181466335] [66050929818412130693077441046821016092265334803401759338667273028766982617779712320, 2954625760580091857825190852826899891492369204494107266499709578 694843084725227571201] [2608077516025604206696668091297297960193250473178344064814349517650835355756863832363680706968, 11666592787753487807500058472831110573381478047673851 4851627220646522719069184329815980490117855]

So minimal x= 441354968

Sorry i mistook and solved $2001x^{2}-y^{2}+1 =0$

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Alpha shows this is not correct. –  Ross Millikan Jan 26 '13 at 23:30
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