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Let be $ f(x,y)= 6xy-3x-2y+1 $

How can I show that the congruence $ f(x,y) \equiv 0 \pmod m $

has a solution in $ \mathbb{Z}^2 $ for any $ m \in \mathbb{N} $?

I know from here that we can factor $f$ as $(3x-1)(2y-1)$ but I don't know how to proceed further.

Erick Wong
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constant94
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1 Answers1

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Hint $ $ Let $\,m = 2^kn,\ \color{#c00}{n\ \rm odd}.\,$ To solve $\,2^k n\mid (3x\!-\!1)(2y\!-\!1)^{\phantom{|^|}}\!\!$ it suffices to solve $\, 2^k\mid 3x\!-\!1\,$ & $\,n\mid 2y\!-\!1^{\phantom{|^|}}\!\!\!,\,$ i.e. $\,x\equiv 3^{-1}\!\pmod{\!2^k}\,$ & $\, y \equiv 2^{-1}\!\pmod{\!n}^{\phantom{|^|}}\!\!;\,$ both exist by $\,(3,2^k)\! =\! 1\! =\! \color{#c00}{(2,n)}^{\phantom{|^|}}\!\!$

Remark $ $ The factorization follows from: completing a square generalizes to completing a product, using the AC-method, viz. $$\begin{eqnarray} &&axy + bx + cy\, =\, d,\ \ a\ne 0\\ \overset{\times\,a}\iff\, &&\!\! (ax+c)(ay+b)\, =\, ad+bc\end{eqnarray}\qquad\qquad$$

J. W. Tanner
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Bill Dubuque
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  • Does AC stand for something, or it’s named after the coefficients a and c? – J. W. Tanner Nov 27 '19 at 23:17
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    @J.W.T Likely the name derives from the coefficient names. I knew the general method long before I learned that some high-school teachers had a name for this special case, so I didn't have a chance to investigate it way back when I was more interested in math history (and had more spare time). – Bill Dubuque Nov 27 '19 at 23:24