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Topic: A quicker way?
Replies: 8   Last Post: Feb 4, 2013 1:44 PM

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quasi

Posts: 10,232
Registered: 7/15/05
Re: A quicker way?
Posted: Feb 4, 2013 12:04 PM
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luttgma@gmail.com wrote:
>quasi wrote:
>> quasi wrote:
>> >luttgma@gmail.com wrote:
>> >>
>> >>Let n^2 = N + a^2, where n, N and a are integers.
>> >>
>> >>Knowing N, it is of course possible to find n by trying
>> >>
>> >> a = 1, 2, 3, 4, etc...
>> >>
>> >>But doesn't exist a quicker way ?

>> >
>> >if you rewrite the equation in the form
>> >
>> > (n - a)(n + a) = N
>> >
>> >then for each pair of integers u,v with u*v = N, you can
>> >
>> >solve the equations
>> >
>> > n - a = u
>> >
>> > n + a = v
>> >
>> >for n and a.

>>
>> Also, since n - a and n + a have the same parity (both even
>> or both odd), you only need to consider pairs of integers u,v
>> with u*v = N for which u,v are both even or both odd.

>
>Thank you, but then you must find u and v, which doesn't
>seem quicker than trying a=1,2,3, etc.


Sometimes it's a lot quicker.

For example, try to find all pairs (n,a) of positive
integers such that n^2 = 2^100 + a^2.

Using the method a = 1,2,3, ... how long do you think it
will take to find even one such pair?

quasi



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