
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.
Please don't ask nonsimple questions as I'm not a mathematician at all.
>What about adding >some little additional restrictions.
What kind of statement is this.
>Both x and y are kth 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

