Re: Solvability of a Diophantine Equation


"Poo-Sung Park" <puzzlist@xxxxxxxxx> wrote in message news:<d48lm9$1fn$1@xxxxxxxxxxxxxxxxxxxxxxxxx>...
> Prove the following:
>
> For any positive integers a and b, the Diophantine equation
>
> x1^2 + x2^2 + a x3^2 + a x4^2 + b x5^2 + b x6^2 = ab
>
> has a solution.

Recall that a number n can be expressed as sum of two squares iff any
prime congruent with -1 modulo 4 dividing n, divides it an even number
of times. Let

A + aB + bC = ab,

where A, B and C fulfill the above condition. If a or b are
representable as sum of two squares, take A=0 and B=b and C=0 or B=0
and C=a as correspond. When neither a nor b are representable, ab is
obviosly representable, then take B=C=0 and A=ab.
.

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