It is well known that $(x^2+y^2)(u^2+v^2)=(xu+yv)^2 + (xv-uy)^2$, thus it suffices to characterize the primes $p$ with $x^2+y^2=p$. Are there any similar composition formulae for $ax^2+by^2$ and $ax^2-by^2$?
Composition formula for $ax^2 \pm by^2$
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number-theory
quadratic-forms
sums-of-squares
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0https://www.artofproblemsolving.com/community/c3046h1359115_representation_of_a_number_the_squares_in_different_ways_2 – 2017-02-21
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0Martin, Gauss composition takes the product of two numbers of the form $\alpha x^2 + \beta y^2$ to a number of the form $u^2 + \alpha \beta v^2.$ In case this is in a non-principal genus, such as $2 x^2 + 3 y^2,$ the product can only be represented by $u^2 + 6 v^2$ by simple congruences, in this case $\pmod 3$ – 2017-02-21
1 Answers
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$$q(ac+nbd)^2+n(ad-qbc)^2=(a^2+nqb^2)(qc^2+nd^2)$$
We must remember that then can be reduced by a common factor.
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0but this is not a composition formula of the required type since the coefficients $a,b$ in $ax^2+by^2$ are not the same on the right-hand side – 2017-02-21
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0@Martin Don't understand. Not satisfied with this form? And what should be? – 2017-02-21
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0In your equation the coefficients $a$ and $b$ of $ax^2+by^2$ on the right-hand side vary ($1$ and $nq$ for the first factor and $q$ and $n$ for the second factor), while I want them to be fixed – 2017-02-21
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0@Martin I want to have only the sum of the squares of the coefficients without? The coefficients should still appear. – 2017-02-21