Number theory proofs regarding gcd's

Question

How would you prove if $ad-bc = 1$, then $(a+c,b+d)=1$

Why don't you ask a counselor? – Dylan Yott Jul 29 '14 at 16:11 — Dylan Yott, Jul 29 '14 at 16:11

score 5 · Accepted Answer · answered Jul 29 '14 at 02:23

5

$$ (a+c)d - (b+d)c = ad - bc = 1 $$

answered Jul 29 '14 at 02:23

David Holden

18,040

score 1 · Answer 2 · edited Apr 13 '17 at 12:21

1

Hint $\ $ The linear map $\,(x,y)\mapsto (ax\!+\!cy,bx\!+\!dy)\,$ has determinant $\,D = ad-bc,\,$ hence, by a simple proof we deduce that $\,\gcd(ax\!+\!cy,bx\!+\!dy)\mid D\gcd(x,y).\,$ Yours is case $\,D = x = y = 1$.

edited Apr 13 '17 at 12:21

Community

1

answered Jul 29 '14 at 02:57

Bill Dubuque

272,048

score 1 · Answer 3 · answered Jul 29 '14 at 05:20

1

You could take the equation;

$$ad-bc=1$$

And see the look of his decision.

$$a=ps+1$$

$$d=ps+p+s+2$$

$$b=ps+p+1$$

$$c=ps+s+1$$

$$.......$$

$$a=2ps+p+s+1$$

$$d=2ps+3p+3s+5$$

$$b=2ps+3p+s+2$$

$$c=2ps+p+3s+2$$

$p,s$ - integers of any sign.

answered Jul 29 '14 at 05:20

individ

4,301

What does this even mean? – Stahl Jul 30 '14 at 17:33
@Stahl this is the formula for solutions of the equation. – individ Jul 30 '14 at 17:41

Number theory proofs regarding gcd's

3 Answers3