Intuitive proof for $(a, b) = (a', b')$.

Question

If $p,q,r,s$ are integers such that $ps - qr = \pm 1$, and $a,b,a',b' $ are integers such that $a' = pa +qb, b' = ra + sb$, prove that $(a, b) = (a', b')$.

Here, the particular value of $ps - qr = \pm 1$ enables us to get integer values.

Can there be an algebraic or intuitive explanation for that?

Have two explanations. My explanation is given for both sequentially. The first explanation is offered as a geometrical / intuitive one.

While the second one uses number theoretic approach, and linear system of equalities.

Request vetting & improvement for the first one

First, needs linear algebra, and orthogonal matrices.

$$\begin{bmatrix} a'\\ b' \\ \end{bmatrix} = \begin{bmatrix} p & q\\ r & s \\ \end{bmatrix} \begin{bmatrix} a\\ b\\ \end{bmatrix} $$

The condition $ps - qr = \pm 1$, states two things about matrix $$\begin{bmatrix} p & q\\ r & s \\ \end{bmatrix} $$ #1. determinant is non-zero; hence invertible matrix.

#2. Also, is an orthogonal matrix.

A function is invertible iff it is bijective. A matrix is only a convenient representation of linear mapping function between two sets, wrt a basis of the domain and co-domain of the function.

#1: So, for linear maps, the determinant of corresponding matrix as non-zero means the mapping is possible in both directions without collapsing of any dimension (row/column).

#2: the area (min. possible contained entity, as order $2$ is the minimum dimension; similarly, for order $3$ matrix, it is volume) is unit as for orthogonal matrix of order $n$ 'that' is given by: $I^n = 1$.

Hence, $(a,b) = (a', b')$.

Think that need some geometrical insight in orthogonal matrices.

Edit :: in case of orthogonal matrices, they practically arise in rotation or reflection, where the determinant is $\pm 1$. Can this explanation be modified to take into account that?

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Second explanation is equivalent, though needs to ignore the matrix representation. But, needs the linear system of equalities to use the Bezout's lemma. This is a number theoretic approach.

$$a' = pa + qb$$ $$b' = ra + sb$$

any common divisor of $a, b$ will divide both $a', b'$, so $(a,b) ≤ (a',b')$.

Solving for $a, b$ in terms of $a', b'$ gives

$$a = (s/(ps-qr))a' -(q/(ps-qr))b'$$

$$b = (-r/(ps-qr))a' +(p/(ps-qr))b'$$

Since $ps-qr =\pm1$, all the coefficients in this system are integers, so $(a',b')$ is a common divisor of $a$ and of $b$, therefore, $(a',b') \le (a,b)$.

Answer 1

As for your first explanation, the equations $$ \begin{bmatrix} a' \\ b' \end{bmatrix} = \begin{bmatrix} p & q \\ r & s \end{bmatrix} \begin{bmatrix} a \\ b \end{bmatrix}$$ imply that any common divisor of $a$ and $b$ is also a common divisor of $a'$ and $b'$. Since $\gcd(a,b)$ thus divides both $a'$ and $b'$ you have $$\gcd(a,b) \le \gcd(a',b').$$

Since $\begin{bmatrix} p & q \\ r & s \end{bmatrix}$ is invertible with integer entries you have also that $$ \begin{bmatrix} a \\ b \end{bmatrix} = \begin{bmatrix} p' & q' \\ r' & s' \end{bmatrix} \begin{bmatrix} a' \\ b' \end{bmatrix}$$ for some integers $p'$, $q'$, $r'$, and $s'$, and thus for the same reason you may conclude $$\gcd(a',b') \le \gcd(a,b).$$

Answer 2

Here's a proof that's short but unintuitive. By multiplying $p, q \text{ and }a' \text{ by } -1$, if necessary, we can assume that $$ps-qr=1.$$

Let's divide $a$ and $b$ on each side by $\gcd(a,b).$ This means that $\gcd(a,b)$ is now $1$ and it only remains to show that $\gcd(a',b')=1.$ There are integers $u$ and $v$ so that $au+bv=1.$ Hence, $$(su-rv)a'+(-qu+pv)b'=au+bv=1,$$ completing the proof.
Or, if you want details: $$(su-rv)(pa+qb)+(-qu+pv)(ra+sb)=$$ $$(psa+qsb-qra-qsb)u+(-pra-qrb+pra+psb)v$$ $$=au+bv=1$$