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Topic: ---- --- ---- conditions for integer solutions
Replies: 15   Last Post: Dec 17, 2010 7:48 PM

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 Ulrich D i e z Posts: 34 Registered: 1/2/10
Re: ---- --- ---- conditions for integer solutions
Posted: Dec 17, 2010 7:48 PM

[ This posting supersedes my posting
news:iegvcp\$3aq\$1@four.albasani.net
wherein I erroneously omitted a closing
parenthese. ]

Deep wrote:

>On Dec 15, 2:07 pm, Ulrich D i e z <eu_angel...@web.de> wrote:
>> Valeri Astanoff wrote:
>> >> On 14 d c, 21:52, quasi <qu...@null.set> wrote:
>>
>> >> > On Tue, 14 Dec 2010 12:25:06 -0800 (PST), Deep <deepk...@yahoo.com>
>> >> > wrote:

>>
>> >> > >Consider (1) below
>>
>> >> > >x^2 -y^2 = ek^2 (1)
>>
>> >> > >Question: Is it possible for (1) to have solutions under the given
>> >> > >conditions?

>>
>> >> > >Conditions: x, y are coprime, each is odd and > 1, e is an even
>> >> > >integer > 1, k is a prime > 3

>>
>> > Is there a solution for any prime k ?
>>
>> Huh? Yes, of course.
>>
>> x^2 - y^2 = ek^2 <-> x^2 = y^2 + ek^2 .
>>
>> If you can write ek^2 with k a prime > 3 as
>> a product 4*s^2*t^2 , s > t > 0, gcd(s,t)=1, s =/= t (mod 2) ,
>> then
>> * e is even > 1
>> * you can use s and t for constructing a
>> primitive pythagorean triple x,y,z with
>> - x = s^2 + t^2
>> y = s^2 - t^2
>> z = 2st
>> - x^2 = y^2 + z^2 = y^2 + ek^2 <-> x^2 - y^2 = ek^2
>> , while such a primitive pythagorean triple fulfills
>> all conditions for x and y.
>>
>> Thus you can e.g. choose:
>> t : = k ; k a prime > 3 ;
>> s : = (k+1) { -> [ s > t > 0 and gcd(s,t)=1 and s=/= t (mod 2) ] }
>> -> [ ek^2 = 4*s^2*t^2 <-> ek^2 = 4*(k+1)^2*k^2 <-> e = 4*(k+1)^2 ]
>> -> [ x = s^2 + t^2 = (k+1)^2 + k^2 ]
>> -> [ y = s^2 - t^2 = (k+1)^2 - k^2 ]
>>
>> [instead of s : = (k+1) you could as well choose
>> s : = (k+m) with gcd(k,m) = 1... ]
>>
>> Example:
>>
>> k :=11 { 11 is a prime > 3 }
>> -> [ e= 4*(k+1)^2 = 4*(11+1)^2 = 576] -> e is an even integer > 1
>> -> [ x = (k+1)^2 + k^2 = (11+1)^2 + 11^2 = 265 = 53*5 ] -> [ x = 1 (mod 2) and x > 1 ]
>> -> [ y = (k+1)^2 - k^2 = (11+1)^2 - 11^2 = 23] -> [ y = 1 (mod 2) and y > 1 ]
>> -> x and y are coprime
>> -> [ x^2 -y^2 = ek^2
>> <->
>> 265^2 - 23^2 = 576*11^2
>> <->
>> 70225 - 529 = 576*121
>> <->
>> 69696 = 69696
>> ->
>> true ]

>Hua! a simple question has generated good responses.

mathematician at all.

What kind of statement is this.

>Both x and y are k-th powers.
>Do such x and y exist?

Huh? Yes, of course. ;-)

I already pointed out that for any prime k > 3 there are solutions
for x^2 - y^2 = ek^2 ;
gcd(x,y) = 1 ;
x >= y > 1 ; [and thus also x > 1 , y > 1]
x = y = e+1 = 1 (mod 2) .

If you have such a solution k, x, y, e with

k = K; x = X; y = Y; e = E

, then

x^2 - y^2 = ek^2

is equivalent to

X^2 - Y^2 = EK^2

, which is equivalent to

(X^2 - Y^2)*(sum{L = 1 .. K}{(X^2)^(K - L)*(Y^2)^(L - 1)}) = EK^2*(sum{L = 1 .. K}{(X^2)^(K - L)*(Y^2)^(L - 1)})

, which is equivalent to

(X^2)^K - (Y^2)^K = EK^2*(sum{L = 1 .. K}{(X^2)^(K - L)*(Y^2)^(L - 1)})

, which is equivalent to

(X^K)^2 - (Y^K)^2 = E*(sum{L = 1 .. K}{(X^2)^(K - L)*(Y^2)^(L - 1)})*K^2

, which with

k = K ;
x = X^K = X^k ;
y = Y^K = Y^k;
e = E*(sum{L = 1 .. K}{(X^2)^(K - L)*(Y^2)^(L - 1)})

is a solution to x^2 - y^2 = ek^2 ;
gcd(x,y) = 1 ;
x >= y > 1 ;
x = y = e+1 = 1 (mod 2) ;
k > 3; k prime
as well while both x and y are powers of k.

Ulrich

Date Subject Author
12/14/10 Deep Deb
12/14/10 quasi
12/14/10 Deep Deb
12/15/10 astanoff
12/15/10 astanoff
12/15/10 Pubkeybreaker
12/15/10 astanoff
12/15/10 Rob Johnson
12/15/10 Ulrich D i e z
12/15/10 Deep Deb
12/17/10 Ulrich D i e z
12/16/10 Ulrich D i e z
12/16/10 alainverghote@gmail.com
12/16/10 Ulrich D i e z
12/16/10 alainverghote@gmail.com
12/15/10 dan73