
Re:    conditions for integer solutions
Posted:
Dec 16, 2010 5:05 AM


On 16 déc, 07:17, Ulrich D i e z <eu_angel...@web.de> wrote: > Ulrich D i e z wrote: > > > > > 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) , > [...] > > Thus you can e.g. choose: > > t : = k ; k a prime > 3 ; > > s : = (k+1) > [...] > > [instead of s : = (k+1) you could as well choose > > s : = (k+m) with gcd(k,m) = 1... ] > > s : = (k+m) with gcd(k,m) = 1, m = 1 (mod 2)... ] > > Sorry, I don't know whether I was thinking at all. > > Ulrich
Bonjour,
It seems to me that all numerical solutions you gave correspond to (e/4+k^2)^2 = (e/4k^2)^2+e*k^2 x^2 = y^2 + e*k^2 add one e=16 , k =7 53^3 = 45^2 +16*7^2 ,
With e/4 even.
Alain

