This question really has me stumped..... (Lipssealed)

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

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- Oct 31st 2010, 09:02 AMDreamer78692show that the equation has only one solution
This question really has me stumped..... (Lipssealed)

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 - Oct 31st 2010, 09:35 AMtonio

The RHS is being expressed as the sum of two squares. This is possible iff all the primes $\displaystyle =3\!\!\pmod 4$ that divide the RHS

appear at an even power....now, we have there $\displaystyle 6=2\cdot 3$ , so....try to complete the argument now.

Tonio - Oct 31st 2010, 11:10 AMDreamer78692
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 - Oct 31st 2010, 11:55 AMtonio
- Oct 31st 2010, 12:04 PMDreamer78692
- Oct 31st 2010, 07:27 PMtonio

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