This question really has me stumped.....
a^2 + b^2 = 6( c^2 + d^2 )
Any clues or hints to start this question...
I think we suppose to use congruences to try an solve this...
Any help greatly appreciated
Sorry Im not quite getting it .....
could you elaborate please...
from what i read....
c^2 + d^2 can be expressed as (2^k)*p1..pk*q1...ql
a^2 + b^2 can be expressed as (2^g)*s1..sm*t1...tn
where p1...pk and s1...sm are primes =1 mod 4
and where q1...ql and t1...tn are primes =3 mod 4 with even power...
then
a^2 + b^2 = 6( c^2 + d^2 )
can be expressed as
(2^g)*s1..sm*t1...tn = 6*(2^k)*p1..pk*q1...ql
am i on the right track... I don't really know where to go from there
Another guy posted this solution is this correct
MathBin.net - Untitled
The only thing I can really think of is what I already told you: it is a very well known result in number theory that the only primes that can
be expressed as the sum of two squares are , and from here we get that a natural
number is expressable as the sum of two squares iff all its prime divisors which are appear in the prime decomposition
of this natural number at even powers.
Thus, for example, can't be written as the sum of two squares, but can...
Tonio