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$\displaystyle 2 \,\,and\,\,those\,\,which\,\,are\,\, =1\!\!\pmod 4$ , and from here we get that a natural
number is expressable as the sum of two squares iff all its prime divisors which are $\displaystyle =3\!\!\pmod 4$ appear in the prime decomposition
of this natural number at even powers.
Thus, for example, $\displaystyle 5\cdot 7\cdot 13$ can't be written as the sum of two squares, but $\displaystyle 5\cdot 7^2\cdot 13$ can...
Tonio