[seqfan] Re: Squares of the form x^2+40y^2 .

[Robert Israel]

Suppose Z is an odd prime.  Any solution of
(1001 Z)^2 = X^2 + 40 Y^2 with X divisible by Z will have Y 
also divisible by Z, and so will be Z times a primitive solution.  A 
solution with X not divisible by Z would require -10 to be a quadratic
residue mod Z.  There are infinitely many primes Z for which -10 is not a 
quadratic residue mod Z.

Cheers,
Robert Israel


On Thu, 22 Jan 2009, Jack Brennen wrote:

> It would seem that it contains an infinite number of multiples of 1001.
>
> Maybe someone can determine what property of the integer Z leads to
> (1001*Z)^2 == X^2 + 40*Y^2 having other solutions than the 13
> primitive solutions multiplied by Z?
>
> The values of Z which do lead to extra solutions include all multiples
> of:
>
> 7, 11, 13, 19, 23, 37, 41, 47, 53, 59, 89, 103, 127, 131, 139, ...
>
> (Which doesn't appear to be a sequence in the OEIS.)
>
> zak seidov wrote:
>> Dear seqfans,
>> Is this sequence finite?
>>
>> %I A000001
>> %S A000001 1001,1463,1729,1771,2002,2093,2717,2849,2926,3003,3059,3157
>> %N A000001 Numbers n with property that n^2 has exactly 13 representations as x^2+40y^2 with positive x, y.
>> %C A000001 Is the sequence finite?
>> %e A000001 n=1001: {x,y}={{169,156},{231,154},{279,152},{371,147},{429,143},{681,116},{749,105},{781,99},{819,91},{871,78},{921,62},{979,33},{999,10}};
>> %e A000001 n=1463: {x,y}={{77,231},{247,228},{627,209},{653,207},{777,196},{787,195},{903,182},{1197,133},{1273,114},{1287,110},{1353,88},{1427,51},{1453,27}}.
>> %Y A000001 A107145 Primes of the form x^2+40y^2.
>> %K A000001 nonn
>> %O A000001 1,1
>> %A A000001 Zak Seidov (zakseidov(AT)yahoo.com), Jan 22 2009
>>
>>
>>
>>
>>
>>
>>
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