show that the equation has only one solution

**Dreamer78692** · October 31st 2010, 09:02 AM

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

**tonio** · October 31st 2010, 09:35 AM

Originally Posted by Dreamer78692

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

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

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

Tonio

**Dreamer78692** · October 31st 2010, 11:10 AM

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

**tonio** · October 31st 2010, 11:55 AM

Originally Posted by Dreamer78692

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

I don't think it is: he only managed to prove, perhaps, that both a,b are multiples of 3...

Tonio

**Dreamer78692** · October 31st 2010, 12:04 PM

Originally Posted by Dreamer78692

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

could you show me how to do it... pretty please

**tonio** · October 31st 2010, 07:27 PM

Originally Posted by Dreamer78692

could you show me how to do it... pretty please

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 $2 \,\,and\,\,those\,\,which\,\,are\,\, =1\!\!\pmod 4$ , and from here we get that a natural

number is expressable as the sum of two squares iff all its prime divisors which are $=3\!\!\pmod 4$ appear in the prime decomposition

of this natural number at even powers.

Thus, for example, $5\cdot 7\cdot 13$ can't be written as the sum of two squares, but $5\cdot 7^2\cdot 13$ can...

Tonio

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