Help: Sequence

A sequence {ana_{n}} of real numbers is defined by a1=1a_{1}=1 and for all integers n1n≥1

an+1=ann²+nn²+n+2an²a_{n+1}=\frac{a_{n}\sqrt{n²+n}}{\sqrt{n²+n+2a_{n}²}}

Compute the sum of all positive integers n<1000n<1000 for which ana_{n} is a rational number.

Note by Ryan Merino
1 month, 3 weeks ago

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Jason Gomez - 1 month, 3 weeks ago

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Hopefully you can continue from here( I have no clue on how to find out when ana_n is rational, or maybe I haven’t put enough thought on that part)

Jason Gomez - 1 month, 3 weeks ago

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Thanks for this big help mate, just need to compute when the numerator, nn, is a perfect and at the same time the denominator, 3n23n-2, is also a perfect square so that ana_{n} is a rational number. Found out that only when n=1,9n=1, 9, and 121121 works, so the sum is 131131. I don't have a clue for when 3n23n-2 is a perfect square when nn is a perfect square so I just tried all perfect squares less than 10001000.

Ryan Merino - 1 month, 3 weeks ago

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@Ryan Merino There can be times where common factors cancel out to give a square

Jason Gomez - 1 month, 3 weeks ago

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@Jason Gomez Thank you for the correction mate. How careless of me, now I have a headache answering this. lol

Ryan Merino - 1 month, 3 weeks ago

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@Jason Gomez I did some simplifications and checked using a program all the 1000 possibilities and tother that the ones mentioned none of the others looked rational(I had to manually check all, so I could have made a mistake)

Jason Gomez - 1 month, 3 weeks ago

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@Jason Gomez Still it was wrong of me to presume that ana_{n} will only be rational iffiff nn and 3n23n-2 are perfect squares. Thanks for all the help mate, really appreciated. :)

Ryan Merino - 1 month, 3 weeks ago

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@Ryan Merino No problem at all

Jason Gomez - 1 month, 3 weeks ago

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@Ryan Merino If you do assume both have to be squares though,a little modular arithmetic taken against four, will lead to the conclusion that nn is of the form 4k+14k+1

Jason Gomez - 1 month, 3 weeks ago

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@Ryan Merino I recently thought of this back again and found out that your hypothesis that both numerator and denominator are both squares is indeed true, the only factor nn and 3n23n-2 can share is two, putting n=k2,2k2n=k^2,2k^2 and 3n2=m2,2m23n-2=m^2,2m^2 for the two conditions where they share no factor and share a two, the equation becomes

3k22=m23k^2-2=m^2 And 3k21=m23k^2-1=m^2

Doing modular arithmetic with respect to three on both sides knowing that m2mod3=0,1m^2 \mod 3 = 0,1, it’s easy to show that the second doesn’t satisfy ( it gives 2mod32 \mod 3 on LHS) and therefore only the case where the numbers share no common factors can be taken

Jason Gomez - 1 month, 2 weeks ago

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@Jason Gomez @Jason Gomez Thank you mate, now it's all clear to me. :)

Ryan Merino - 1 month, 2 weeks ago

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