Problem 137

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Killa
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Joined: Wed Jan 26, 2011 7:38 pm

Problem 137

Post by Killa »

I found out that problem can be solved by finding first 15 integer terms of sequence sqrt(*snip*). I've tried to get then in many different ways, using c++ or mathematica, and always my algorithms are too slow. I could use some hint to find those, or some another idea to solve this problem.

Killa
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stijn263
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Re: Problem 137

Post by stijn263 »

Can you perhaps spot a pattern among the first few values of n for which sqrt(*snip*) is integer?

Have fun solving!
Killa
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Re: Problem 137

Post by Killa »

To be honest i've seen this integers first time in my life (and im not new to math; )).

Friend helped me to solve it, so problem is already gone: ).
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hk
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Re: Problem 137

Post by hk »

If you're not new to math you've probably heard of Pell's equation.
In this case you're asked to find solutions to a Pell-like quadratic diophantine equation
So here are two keywords you can use to solve this kind of problem.
This kind of questions are not uncommon here.
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Killa
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Re: Problem 137

Post by Killa »

Yeah, i've been thinking some about problem 140, kinda similiar to 137 and i found the connection with pell equasions, in some days i'll certainly find the answers and learn some new things.

Thanks for the hints: )
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GenePeer
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Re: Problem 137

Post by GenePeer »

Can someone who solved the problem send me a PM explaining hyperdex's solution. In particular, I don't understand why he does this:
hyperdex wrote:if the numerator is (~) mod (~), we check to see if it leads to a solution. If the numerator is (~) mod (~), we check to see if (~) the numerator leads to a solution. (Since we are looking for (~), we don't need to check any further multiples of the numerator.)
Why don't we need to check further multiples? His post is the twelfth on the first page. Thank you.
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Lord_Farin
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Re: Problem 137

Post by Lord_Farin »

PM sent.
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RishadanPort
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Re: Problem 137

Post by RishadanPort »

I think there is some mistake on the wording of this question.

The question does say that we are considering only values of x such that Af(x) is a positive integer.

However, there are solutions for negative values of x.

IE x = -2 and x = 1/2 both result in Af(x) = 2


However for the sake of the problem, I guess that Af(x) = 2 is rational for both values of X, positive and negative
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RishadanPort
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Re: Problem 137

Post by RishadanPort »

I solved this problem...

An amazing problem, props to you guys
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DJohn
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Re: Problem 137

Post by DJohn »

RishadanPort wrote: Mon Aug 05, 2019 11:22 pm IE x = -2 and x = 1/2 both result in Af(x) = 2
$A_F(-2) \neq 2$. It doesn't converge.
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RishadanPort
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Re: Problem 137

Post by RishadanPort »

Oh!

Sorry then. thanks for the reply.
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n2sz
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Problem 137

Post by n2sz »

This is to confirm (I am pretty sure this is true) that the first nugget is the 1/2 explained in the problem statement. I have now found (bug in code) that was leading me to question whether this one wasn’t counted in the tally, but it is indeed nugget #1. Maybe that will help another lost soul…

Now off to find 11 thru 15…
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