Solving simple congruences by hand

Question

When I am faced with a simple linear congruence such as $$9x \equiv 7 \pmod{13}$$ and I am working without any calculating aid handy, I tend to do something like the following:

"Notice" that adding $13$ on the right and subtracting $13x$ on the left gives: $$-4x \equiv 20 \pmod{13}$$

so that $$x \equiv -5 \equiv 8 \pmod{13}.$$

Clearly this process works and is easy to justify (apart from not having an algorithm for "noticing"), but my question is this: I have a vague recollection of reading somewhere this sort of process was the preferred method of C. F. Gauss, but I cannot find any evidence for this now, so does anyone know anything about this, or could provide a reference? (Or have I just imagined it all?)

I would also be interested to hear if anyone else does anything similar.

Answer 1

By Gauss' algorithm, scale $\rm\:\color{#C00}{\frac{A}B} \to \frac{AN}{BN}\: $ by the least $\rm\,N\,$ so that $\rm\, BN > 13,\, $ reduce mod $13,\,$ iterate.

$$\rm\displaystyle \ mod\ 13\!:\ \color{#C00}{\frac{7}9} \,\equiv\, \frac{14}{18}\, \equiv\, \color{#C00}{\frac{1}5}\,\equiv\, \frac{3}{15}\,\equiv\, \color{#C00}{\frac{3}2} \,\equiv\, \frac{21}{14} \,\equiv\, \color{#C00}{\frac{8}1}$$

Denominators of the $\color{#c00}{\rm reduced}$ fractions decrease $\,\color{#C00}{ 9 > 5 > 2> \ldots}\,$ so reach $\color{#C00}{1.}$

Or, simpler, allowing negative residues $\displaystyle\ \ \frac{7}9\,\equiv\, \frac{7}{\!-4\!\ \,}\,\equiv\,\frac{21}{\!\!-12\ \ \ \!\!}\,\equiv\, \frac{8}1$

This optimization of using balanced residues $0,\pm 1, \pm 2.\ldots$ works for modular arithmetic in general. Here we can also optimize by (sometimes) cancelling obvious common factors, or by pulling out obvious factors of denominators, etc. For example

$$\frac{7}9\,\equiv\, \frac{\!-6\,}{\!-4\,}\,\equiv\frac{\!-3\,}{\!-2\,}\,\equiv\frac{10}{\!-2\,}\,\equiv\,-5$$

$$\frac{7}9\,\equiv\,\frac{\!-1\cdot 6}{\ \ 3\cdot 3}\,\equiv\,\frac{\!\,12\cdot 6\!}{\ \ \,3\cdot 3}\,\equiv\, 4\cdot 2$$

As you did, we can check if the quotient $\rm\,a/b\equiv (a\pm\!13\,i)/(b\pm\!13\,j)\,$ is exact for small $\rm\,i,j,\,$ e.g.

$$ \frac{1}7\,\equiv \frac{\!-12}{-6}\,\equiv\, 2;\ \ \ \frac{5}7\,\equiv\,\frac{18}{\!-6\!\,}\,\equiv -3$$

When working with smaller numbers there is a higher probability of such optimizations being applicable (the law of small numbers), so it's well-worth looking for such in manual calculations.

Note $ $ Gauss' algorithm is my name for a special case of the Euclidean algorithm that's implicit in Gauss' Disq. Arith. I don't recall if Gauss explicitly used this algorithm. Follow said link for more.

Answer 2

When the prime is a reasonably small one I'd rather find directly the inverse: $$9^{-1}=\frac{1}{9}=3\pmod {13}\Longrightarrow 9x=7\Longrightarrow x=7\cdot 9^{-1}=7\cdot 3= 21=8\pmod {13}$$ But...I try Gauss's method when the prime is big and/or evaluating inverses is messy.