Help: Sequence

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

$a_{n+1}=\frac{a_{n}\sqrt{n²+n}}{\sqrt{n²+n+2a_{n}²}}$

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

#Algebra

Easy Math Editor

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Comments

Jason Gomez - 3 months, 2 weeks ago

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

Jason Gomez - 3 months, 2 weeks ago

Thanks for this big help mate, just need to compute when the numerator, $n$ , is a perfect and at the same time the denominator, $3n-2$ , is also a perfect square so that $a_{n}$ is a rational number. Found out that only when $n=1, 9$ , and $121$ works, so the sum is $131$ . I don't have a clue for when $3n-2$ is a perfect square when $n$ is a perfect square so I just tried all perfect squares less than $1000$ .

Ryan Merino - 3 months, 2 weeks ago

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

Jason Gomez - 3 months, 2 weeks ago

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

Ryan Merino - 3 months, 2 weeks ago

@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 - 3 months, 2 weeks ago

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

Ryan Merino - 3 months, 2 weeks ago

@Ryan Merino – No problem at all

Jason Gomez - 3 months, 2 weeks ago

@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 $n$ is of the form $4k+1$

Jason Gomez - 3 months, 2 weeks ago

@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 $n$ and $3n-2$ can share is two, putting $n=k^2,2k^2$ and $3n-2=m^2,2m^2$ for the two conditions where they share no factor and share a two, the equation becomes

$3k^2-2=m^2$ And $3k^2-1=m^2$

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

Jason Gomez - 3 months, 1 week ago

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

Ryan Merino - 3 months, 1 week ago