(In reply to
re: solution by JayDeeKay)
good catch, missed the part about infinite solutions.
A=B=2^p
then we have
A^n+B^n=
2*[2^p]^n
2*2^(pn)
2^(pn+1)
so for this to be a perfect (n1) power we need
pn+1 = 0 mod (n1)
but n = 1 mod (n1) thus this reduces to
p+1 = 0 mod (n1)
p = 1 mod (n1)
p = n2 mod (n1)
p=(n1)k+n2
p=knk+n2
p=(k+1)nk2
thus for any given N we have a family of solutions:
p=(k+1)nk2
A=B=2^p
C=2^[(pn+1)/(n1)]
for all k>=0

Posted by Daniel
on 20150512 04:45:21 