# [seqfan] Re: Fixing A175155 Numbers n satisfying n^2 + 1 = x^2 y^3

Jack Brennen jfb at brennen.net
Fri Nov 16 17:57:24 CET 2012

```Note that stating that a number is of the form x^2 y^3 is equivalent
to stating that the number is powerful.

So the sequence is also described as numbers n such that n^2+1 is
powerful.  But shouldn't the sequence begin with 0?

On 11/16/2012 8:23 AM, Charles Greathouse wrote:
> Good catch, Georgi.
>
> I wouldn't add a new sequence, just correct this one. (Unless someone feels
> that this would be worthwhile?)
>
> Charles Greathouse
> Analyst/Programmer
> Case Western Reserve University
>
>
> On Fri, Nov 16, 2012 at 10:56 AM, Georgi Guninski <guninski at guninski.com>wrote:
>
>> A175155 Numbers n satisfying n^2 + 1 = x^2 y^3
>>
>> I am not sure this is entirely correct:
>>> This sequence is infinite. The fundamental solution of n^2 + 1 = x^2 y^3
>> is (n,x,y) = (682,61,5), that mean the Pellian equation n^2 - 125x^2 = -1
>> has the solution (n,x) = (682,61) =(n(1),x(1)). Then, this Pellian equation
>> admit an infinity solutions (n(2k+1),x(2k+1))
>>
>> This indeed is a family of solutions giving the smallest one,
>> but there are infinitely many other solutions arising from
>> x^2 - k^3 y^2 = -1
>>
>> In particular n=1459639851109444 is missing from the sequence.
>> n^2 + 1 = 17^3 * 79153^2 * 263090369^2
>>
>> I suggest:
>> 1. Adding the missing term and other low hanging fruit from pell eqs
>> 2. Indicating that terms might be missing (the sequence contains a
>> 22 digit number and I suppose it is infeasible to find all terms up to
>> it)
>>
>> Should I submit another sequence that currently numerically coincides with
>> A175155?
>>
>> Solution x to x^2 - 125 y^2 = -1
>>
>>
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