Number theory proofs regarding gcd's

Question

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

Answer 1

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

Answer 2

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 $\,D = x = y = 1$.

Answer 3

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.